Chapter 12 Electrochemistry

The oxidation number of each group VIIA element in a compound is ________, except when combined with an element of higher electronegativity

-1

In most compounds, oxygen has an oxidation number of Exceptions In peroxides, oxygen has a charge of _______ Compounds where oxygen is not the ________ electronegative

-2 -1 most

Free energy change for a redox reaction

-nFE; if cell voltage is positive, reaction is spontaneous

The sum of the oxidation numbers in a neutral compound is always

0

What is the oxidation state of a free element?

0

Determine the K for this reaction Fe(s) + Cd2+ (aq) --> Fe2+ + Cd(s) Reduction values: Cd: -.403 Fe: -.44

17.8

The two electrodes are connected to each other by what?

A conductive materials such as copper wire

What is a combustion reaction?

A fuel is mixed with an oxidant to form CO2 and H2O

What is standard hydrogen electrode?

A half cell in which a solution of H+ is in equilibrium with H2 gas on the surface of a platinum electrode under standard conditions

What is energy density?

A measure of a battery's ability to produce power as a function of its weight.

What is an oxidizing agent?

A substance that oxidizes other molecules by reducing itself.

What is a reducing agent?

A substance that reduces another substance by oxidizing itself

In a Galvanic cell, the Zn2+ ions that remain in solution in the zinc half cell attract _______ ions from the salt bridge, allow K+ ions in the salt bridge to be attracted to the copper half of the cell

NO3-

Electrolytic cells are __________

non-spontaneous

In electrolytic cells, the anode is

positive

Oxidizing agents have the most __________ reduction potential

positive

The cathode in a galvanic cell is

positive

The stronger oxidizing agents have a more ________ reduction potential

positive

What does a negative cell voltage mean?

reaction is nonspontaneous

What does a positive cell voltage mean?

reaction is spontaneous

In a typical electrolytic cell, sodium is __________ and chlorine is __________

reduced oxidized

A typical example of an electrolytic cell is one used to form ________ metal and _________ gas from molten ________

soidum chlorine NaCl

What is the anode of a nickel cadmium battery?

solid cadmium

All galvanic cells are

spontaneous

The more positive the reduction potential, the _______ the reactant is an oxidizing agent and the _______ the product is a reducing agent

stronger weaker

What is the standard reduction potential?

tendency of a species to gain electrons to be reduced.

What is Faraday's Law of Electrolysis?

the amount of chemical change is proportional to the amount of electricity that flows through the cell

What is standard reduction potential?

the measure of affinity a compound has for electrons

What is the oxidation state of a monoatomic ion?

the same as its charge Na+ = 1 Cu2+ = 2

What are combination reactions?

two or more substances react to form one product H2(g) + F2 (g) --> 2 HF (aq) Half reactions: H2 --> 2H+ 2 e- F2 + 2e- --> 2F-

What is a concentration cell?

type of galvanic cell in which the only difference is concentrations.

The more negative the reduction potential, the _________ the reactant is as an oxidizing agent, and the ________ the product is as a reducing agent

weaker stronger

Group IA elements have an oxidation number of

+1

Oxidation number of Cl in HOCl

+1

Oxidation number of hydrogen in HCl

+1

The oxidation number of hydrogen is usually ________ The oxidation number when bonded to less electronegative elements is ________

+1 -1

Group IIA elements in a compound have an oxidation number of

+2

Oxidation number of Cl in HCl

-1

Oxidation number of H in NaH

-1

What is the reduction half reaction in a nickel cadmium battery?

2 NiO(OH) (s) + Cd (s) + H2O --> 2 Ni(OH) 2 (s) + Cd(OH)2 (s)

Consider the following table: ReactionE° F2(g) + 2e- 2 F-+2.87 volts Cl2(g) + 2e- 2 Cl-+1.36 volts Which of the following best describes the following reaction? F2(g) + 2 Cl- 2 F- + Cl2(g) A. Spontaneous B. Metal-halogen exchange C. Nonspontaneous D. Neutral

A

Mn3+ + e- --> Mn2+ Fe3+ + e- --> Fe2+ According to the reduction potentials given above, Mn3+ is a much stronger oxidizing agent than Fe3+. However, one would expect that Fe3+ with its larger nuclear charge and slightly smaller size should gain electrons more easily than Mn3+. Which one of the following best accounts for this behavior? A. The Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+gives the relatively stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a stable 3d5 configuration to a less stable 3d6 configuration. B. Fe3+ has a higher ionization energy than Mn3+, so it attracts electrons more strongly than Mn3+. C. The Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+gives the less stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a 3d5 configuration to a 3d6 configuration, which is more stable since it is closer to a filled 3d10 configuration. D. Mn3+ is more electropositive than Fe3+, so it should attract electrons more strongly.

A

What is the oxidation state of iron in FeO42-? A. +6 B. +2 C. +3 D. +8

A

In the electrochemical cell describes by the following cell diagram, what reaction occurs at the anode? Zn (s) | Zn2+ (aq) || Cl- (aq) | Cl2 (g) A. Zn —> Zn2+ + 2e- B. Zn2+ + 2e- —> Zn C. 2 Cl- —> Cl2 + 2e- D. Cl2 + 2e- —> 2 Cl-

A (oxidation always occurs at the anode)

In an electrolytic cell containing molten MgCl2(l): A. Mg2+ is reduced at the cathode, and Cl- is oxidized at the anode. B. Mg2+ is oxidized at the anode, and Cl- is reduced at the cathode. C. Mg2+ is reduced at the anode, and Cl- is oxidized at the cathode. D. Mg2+ is oxidized at the cathode, and Cl- is reduced at the anode.

A;

In an electrolytic cell, the electrochemical reaction is: A. nonspontaneous, with oxidation occurring at anode. B. nonspontaneous, with oxidation occurring at the cathode. C. spontaneous, with oxidation occurring at the anode. D. spontaneous, with oxidation occurring at the cathode.

A;

In the net reaction of the pyruvate dehydrogenase complex displayed below, NAD+ functions as a(n): pyruvate + HSCoA + NAD+ → acetyl-CoA + NADH + H+ + CO2 A. oxidizing agent. B. reducing agent. C. catalyst. D. enzyme.

A;

Which one of the following is NOT a characteristic of the anodic electrode in an electrochemical cell? A. It is the site of reduction. B. Anions from the salt bridge migrate toward it. C. It is the location where free electrons are formed. D. It is often consumed during the course of the reaction.

A;

In the span of two hundred minutes, 30 grams of copper is plated on a sample of iron with a constant current of 3.8 A. How many moles of electrons are transferred per mole of plated copper? Note: F = 96,500 C/mol e- A. 1 mole B. 2 moles C. 3 moles D. 4 moles

A; 30 g Cu x (1 mol Cu/64 g Cu) x (x moles e-/1 mole Cu) (96,500 C/1 mole e-) divided by 12,000 s = 3.8 A x moles e- = (3.8A x 12,000s x 1 mole Cu) / 30g Cu (1 mole Cu/64gCu) (96,500 C/1 mole e-) = 4.5 x 10^4/5x10^4 = 1 mol e-

Which of the following best approximates the value of delta G degree for this reaction? 2 Al + 3Cu2+ —> 2 Al3+ + 3 Cu A. -(12)(96,500) J B. -(6)(96,500) J C. +(6)(96,500) J D. +(12)(96,500) J

A; The half reactions are 2(Al —> Al3+ + 3e-) E = +1.67 V 3 (Cu2+ + 2e-) —> Cu E = .34V So the overall cell voltage is about 2V. Because the number of electrons transferred is n = 2 x 3 = 6, the equation delta G degree = -nFE Delta G degree = -(6)(96,500)(2) = -(12)(96,500)J

A galvanic cell is set to operate at standard conditions. If one electrode is made of magnesium and the other is made of copper, then the magnesium electrode will serve as the: (Mg2+ + 2e- —> Mg (-2.36) (Cu2+ + 2e- —> .34) A. Anode and be the site of oxidation B. Anode and be the site of reduction C. Cathode and be the site of oxidation D. Cathode and be the site of reduction

A; Anode always serves as the site of oxidation and cathode always serves as the site of reduction. Eliminate choices B and C From the data, we see that the reduction of Mg2+ is nonspontaneous, whereas the reduction of Cu2+ in spontaneous. Therefore, the copper electrode will serve as the cathode and be the site of reduction, and the magnesium electrode will serve as the anode and be the site of oxidation

Based on the following data: Li+(aq) + e- Li(s) E° = -2.71 V Na+(aq) + e- Na(s) E° = -3.05 V which one of the following must be true? A. Sodium metal is more easily oxidized than lithium metal. B. Lithium metal is more easily oxidized than sodium metal. C. Lithium metal is more easily reduced than sodium metal. D. Both lithium and sodium ions are spontaneously reduced by H2.

A; Based on the following data sodium is more easily oxidized than lithium. Li+(aq) + e- Li(s) E° = -2.71 V Na+(aq) + e- Na(s) E° = -3.05 V Since the reduction potentials for Li+(aq) and Na+(aq) are both negative, neither process is spontaneous, so "both sodium and lithium are spontaneously reduced" is wrong. Flipping each of the reactions listed in the table reveals that Na(s) is more easily oxidized than Li(s), since 3.05 V > 2.71 V. Neither ion is spontaneously reduced by H2, since the oxidation of H2 is generally defined as 0, and the overall redox equations would have a negative voltage. The redox potentials given do not give any insight into the further reduction of Na or Li metal.

A drained battery is placed onto its charger causing its normal oxidation-reduction reaction to proceed in reverse. Which of the following best describes the changes in the cell? A. Both electrodes maintain the same charge following reversal of the reaction. B. Electron flow occurs from the cathode to the anode. C. The reverse reaction occurs spontaneously. Your Answer D. The cathode becomes the site of oxidation.

A; Both electrodes maintain the same charge following reversal of the reaction upon placing the battery on its charger. When a discharged battery is attached to a voltage source, the reaction proceeds in reverse. This results in the reversal of the anode and cathode (the cathode is still the site of reduction) and the flow of electrons from the new anode to the new cathode. This reversal of the normally spontaneous reaction of a battery requires an attached voltage source, so it would be nonspontaneous, or an electrolytic cell. In an electrolytic cell, the cathode has a negative charge while the anode has a positive charge. Given that the electrodes changed from anode to cathode (and cathode to anode), the electrodes maintain the same charge following reversal of the cell. This answer can always be arrived at using process of elimination as the remaining answer choices are not true for any electrolytic cell.

Which of the following is LEAST likely to disrupt the function of a single-chamber electrolytic cell designed for electroplating silver? A. Severing the salt bridge B. Removing the voltage source C. Exhausting silver cation D. Drop in reaction vessel fluid level below the anode

A; Electrolytic cells do not require two separate reaction vessels: this eliminates the need for a salt bridge, making choice A the best answer. Electroplating is a nonspontaneous process, so without a voltage source the reaction cannot occur (eliminating choice B). With no starting material (Ag+), the reaction cannot occur (eliminate choice C), and with no contact of the solution to the source of electrons, there will be no closed circuit so electrons cannot flow (eliminate choice D).

Based on the following half-reaction potentials, Sn4+(aq) + 2e- Sn2+(aq) E° = -0.14 V Ag+(aq) + e- Ag(s) E° = +0.80 V Cr3+(aq) + 3e- Cr(s) E° = -0.74 V Fe2+(aq) + 2e- Fe(s) E° = -0.44 V Which of the following is the strongest reducing agent? A. Fe2+(aq) B. Cr(s) C. Ag(s) D. Sn2+(aq)

B

In the construction of a "lemon battery", a galvanized nail coated in zinc and a penny are inserted on opposite ends of a lemon. Each metal is then attached to a voltmeter by a wire and a reading of -0.85 V is obtained. Which of the following best describes this reading? A. The reaction is spontaneous but the leads from the voltmeter to the anode and cathode have been reversed. B. The reaction is nonspontaneous and will proceed without a power source. C. The reaction is spontaneous but requires a salt bridge to complete the circuit. D. The reaction is nonspontaneous and requires a power source to proceed.

A; In the construction of a "lemon battery", a galvanized nail coated in zinc and a penny are inserted on opposite ends of a lemon. Each metal is then attached to a voltmeter by a wire and a reading of -0.85 V is obtained. This is because the reaction is spontaneous but the leads from the voltmeter to the anode and cathode have been reversed. The "lemon battery" must proceed via a spontaneous reaction or there would be no flow of current. An incomplete circuit would result in no electron flow. Therefore obtaining a negative reading must be due to a reversal of the leads.

MnO4-(aq) + 8 H+(aq) + 5 Fe2+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) It takes 35 mL of a 0.093 M solution of potassium permanganate to react completely with 10 mL of an aqueous solution of iron(II) in a redox titration, according to the reaction above. Which of the following represents the molarity of the iron(II) solution? A. (5)(35)(0.093)/(10) B. (10)/(5)(0.093)(35) C. (35)(0.093)/(10) D. (5)(10)/(35)(0.093)

A; MnO4-(aq) + 8 H+(aq) + 5 Fe2+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) It takes 35 mL of a 0.093 M solution of potassium permanganate to react completely with 10 mL of an aqueous solution of iron(II) in a redox titration, according to the reaction above. To find the molarity, we need to divide the number of moles of Fe2+ by the volume of the Fe2+ solution. Therefore, 10 should be in the denominator (choices B and D can be eliminated).Since the denominator is in mL, the numerator should be in mmol to get a final answer in M. Multiply volume by concentration to get (35)(0.093) mmol of MnO4- used in the titration. Looking at the coefficients in the balanced reaction, this corresponds to (5)(35)(0.093) mmol of Fe2+. Putting both components together gives (5)(35)(0.093)/(10), or choice A.

Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) If a galvanic cell with the overall equation shown above was connected to a voltage source and forced to run in reverse, which of the following would be the cathode? A. Cu(s) B. Ag+(aq) C. Cu2+(aq) D. Ag(s)

A; Neither of the ions in solution can serve as the cathode, so choices B and C can be eliminated. The cathode is the site of reduction (red cat). In the reverse reaction, silver is being oxidized to silver ion while copper ion is being reduced to copper solid. This makes Cu(s) the cathode (choice A is correct) and Ag(s) the anode.

If the equation below represents an overall reaction in an electrochemical cell, which of the following materials likely constitutes the anode? 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) A. Cu(s) B. CuNO3(s) C. NO3-(aq) D. H+(aq)

A; The anode is the site of oxidation (an ox) and in the above reaction, copper is oxidized (loses an electron) from Cu(s) to Cu2+(aq) making Cu(s) the anode (choice A is correct). While nitrate is the species being reduced to nitrogen monoxide, this would be taking place at the cathode (eliminate choice C). Hydrogen is neither oxidized nor reduced in the reaction (eliminate choice D) and CuNO3 is the mono-oxidized product of the Cu anode (eliminate choice B).

Which of the following pairs will generate an electrolytic cell with the least stable products? Al3+ + 3 e− → Al(s) E° = -1.68 V Cd2+ + 2 e− → Cd(s) E° = -0.40 V Ag+ + e− → Ag(s) E° = 0.80 V Br2(l) + 2 e− → 2Br− E° = 1.08 V A. Al3+ and Ag(s) B. Br2(l) and Al(s) C. Cd2+ and Br- D. Ag+ and Cd(s)

A; The electrolytic cell with the most negative electrical potential will have the least stable products (choice A is correct). To determine the overall potential of any given cell, the reduction and oxidation half-reactions are added. To find the oxidation half-reactions, the reduction half-reaction in question can be reversed and its potential reversed in sign. In this instance to find the oxidation half-reaction, the reduction of silver above can be reversed to Ag(s) → Ag+ + e− E° = -0.80 V which can then be add to the reduction half-reaction Al3+ + 3 e− → Al(s) E° = -1.68 V. Note that while the oxidation half-reaction is multiplied 3 to appropriately balance the reaction, this does not impact its electrical potential. The overall electrical potential can be found by adding the two half-reaction potentials: -1.68 V + (-0.80 V) = -2.49 V.

What is a complete ionic equation?

Accounts for all ions present in a reaction

Identify the oxidation states of the relevant atoms, the oxidizing agent, and the reducing agent. Al + BPO4 —> B + AlPO4

Al = 0 PO4 = -3 B = +3 B = 0 Al = +3 PO4 = -3 Al is oxidized (reducing agent) B is reduced (oxidizing agent)

What surrounds the electrodes in a galvanic cell?

An aqueous solution composed of cations and anions.

In a Daniell cell, what is the anode and cathode?

Anode: zinc in aqueous ZnSO4 Cathode: copper in CuSO4

What are oxidation numbers?

Assigned to atoms in order to keep track of the redistribution of electrons during chemical reactions.

Al3+ + 3e- Al E° = -1.67 Au3+ + 3e- Au E° = +1.50 Based on the half-reactions listed above, which of the following has the strongest activity? A. Al3+ as an oxidizing agent B. Al as a reducing agent C. Au3+ as a reducing agent D. Au as a reducing agent

B

In the absence of a salt bridge, charge separation develops. The anode develops a positive charge and the cathode develops a negative charge, quickly halting the flow of electrons. In this state, the battery resembles: A. A resistor B. A capacitor C. A transformer D. An inductor

B

In the formation of CO2 through the reaction of O2 and C(s), graphite acts as a(n): A. oxidizing agent. B. reducing agent. C. catalyst. D. intermediate.

B

The oxidation states of sulfur in H2SO4 and H2SO3 are, respectively: A. +2 and +4. B. +6 and +4. C. +4 and +6. D. +4 and +2.

B

Which of the following can be deduced from the following reaction? 2 Fe2+ + H2O2 + 2 H+ → 2 Fe3+ + 2 H2O E°= 1.0 V A. The reaction is nonspontaneous. B. At equilibrium, the products are favored. C. The products are less stable than the reactants. D. The oxidation half-reaction is more spontaneous than the reduction half-reaction.

B

Which of the following species is the oxidizing agent in the following redox reaction? Zn + Cu2+ Zn2+ + Cu A. Zn B. Cu2+ C. Cu D. Zn2+

B

With the following standard redox potentials in mind, reaction of which of the following pairs will proceed with the greatest spontaneity? Al3+ + 3 e− → Al(s) E° = -1.68 V Fe2+ + 2 e− → Fe(s) E° = -0.44 V Cu+ + e− → Cu(s) E° = 0.52 V Ag+ + e− → Ag(s) E° = 0.80 V A. Al3+ and Ag(s) B. Fe2+ and Al(s) C. Ag+ and Cu(s) D. Cu+ and Ag(s)

B

Addition of which of the following to Cd(s) would result in no electrochemical reaction between the two? Fe2+ + 2 e− → Fe(s) E° = -0.44 V Cd2+ + 2 − → Cd() E° = -0.40 V Sn2+ + 2 e− → Sn() E° = -0.14 V Ag+ + e− → Ag() E° = 0.799 V Br2(l) + 2 e− → 2 Br− E° = 1.078 V A. Sn2+(aq) B. Fe2+(aq) C.Ag+(aq) D. Br2(l)

B;

The concentration of Cu2+ in a sample can be determined by titration with iodide, as described in the equation below. 2 Cu2+ + 4 I− → 2 CuI + I2 If an endpoint, determined by the iodide-starch test, is achieved after titration with 2.2 g of KI, how many moles of electrons have been transferred? A. 0.001 moles B. 0.007 moles C. 0.014 moles D. 0.026 moles

B; 2.2 g of KI (MW = 166 g/mol) represents: 2.2/166 = 22/16 x (10−2) = 3/2 x (10−2) ≈ 0.015 moles I− According to the stoichiometry of the reaction, 2 of the 4 total I− ions are oxidized to I0, while 2 remain I−, meaning ~0.007 moles of electrons have been transferred.

Which of the following are advantageous properties for a titrant to have in a redox titration? I. The net electrochemical reaction must have positive electric potential. II. The titrant must be able to be standardized or exist in a pure solid form. III. The titrant must have different colors for its reduced and oxidized forms. A. II and III only B. I and II only C. I only D. II only

B; Advantageous properties for a titrant in a redox titration are that the net electrochemical reaction must have positive electric potential, and the titrant must be able to be standardized or exist in a pure solid form. This is a Roman numeral question, so assess each numbered item as true or false, and eliminate answer choices with each decision. The reaction must be spontaneous for the titration to occur, meaning ΔG must be negative and Eo must be positive. Item I is therefore true, and the "II only" and "II and III only" choices are therefore incorrect. All titrants for all titrations need to have a known concentration with a high degree of certainty. Thus a solution must be able to be made from a dry, pure sample of known mass and volume, or a solution must be able to be standardized with another compound which does have a known concentration. Item II is true as well, so the "I only" choice is incorrect. While there is no reason to even read Item III from a strategy perspective, although having a titrant that has two different colors for its oxidized and reduced forms within the titration is helpful, it is not required. All redox titrations have the advantage that by measuring the potential of the solution during the titration, one can determine the equivalence point of the titration by using graphical methods of analysis.

H2 is removed from an electrochemical cell in which metallic magnesium is oxidized, and H+ is reduced. At STP the volume of gas is measured to be 11 L. If the reaction has proceeded for 2 hours at constant current, what was the value, in amps, of the current? (F = 96,500 C/mol) A. 19.9 A B. 13.4 A C. 3.2 A D. 6.5 A

B; H2 is removed from an electrochemical cell in which metallic magnesium is oxidized, and H+ is reduced. At STP the volume of gas is measured to be 11 L. If the reaction has proceeded for 2 hours at constant current, the value of the current is 13.4 A. (F = 96,500 C/mol) At STP 1 mole of gas occupies 22.4 L. Thus, 11 L accounts for half a mole of H2. Since 2 electrons are required to produce each molecule of H2, the production of 0.5 moles of gas requires a total of 1 mole of e-. Current can then be obtained:

In a fully charged lead-acid battery, metallic Pb is one of the two electrode materials. Which of the following redox reactions is possible during recharge of a spent lead-acid battery? A. Pb → PbSO4 + 2e− B. PbSO4 + 2e− → Pb C. Pb → PbO2 + 4e− D. Pb(H2O)4 + 2e− → Pb

B; If Pb0 is an electrode material in the fully charged battery, then it will be oxidized during discharge since anionic Pb is not feasible. The recharge reaction must then be the reduction of a cationic Pb species to Pb0 (choice B is the correct answer). Eliminate choices A and C as they do not produce Pb0, but describe the oxidation of Pb0, which would take place during discharge. Choice D is not an electrically balanced equation as both lead species are in the 0 oxidation state.

Consider the following reduction reactions and their standard potentials: Cu2+ + 2e- ? Cu(s)E° = +0.337 V Zn2+ + 2e- ? Zn(s)E° = -0.763 V If a piece of Zn metal is dropped into a CuSO4(aq) solution, and a piece of Cu metal is dropped into a ZnSO4(aq) solution, then: A. the zinc metal will begin to dissolve in the CuSO4 solution since Zn(s) is a stronger oxidizing agent than Cu(s). B. the zinc metal will begin to dissolve in CuSO4 solution since Zn(s) is a stronger reducing agent than Cu(s). C. the copper metal will begin to dissolve in the ZnSO4 solution since Cu(s) is a stronger oxidizing agent than Zn(s). D. the copper metal will begin to dissolve in the ZnSO4 solution since Zn(s) is a stronger reducing agent than Cu(s).

B; If a piece of Zn metal is dropped into a CuSO4(aq) solution, and a piece of Cu metal is dropped into a ZnSO4(aq) solution, then the zinc metal will begin to dissolve in CuSO4 solution since Zn(s) is a stronger reducing agent than Cu(s). The voltages given for these redox half-reactions indicate that Cu2+ likes to be reduced, but Zn2+doesn't. Therefore, Zn metal will be oxidized in CuSO4 solution—so Zn metal is the reducing agent—and Cu would precipitate out. The net ionic equation is Zn(s) + Cu2+ → Cu(s) + Zn2+.

An intern in a chemistry laboratory decided to measure the voltage of a galvanic cell and, while constructing the cell, selected a lead chloride salt bridge. Unfortunately the measured voltage of the cell was dramatically lower than anticipated and an unexpected solid began forming in one of the beakers. What caused this deviation from the expected voltage? A. Side reactions due to the oxidation of lead B. Side reactions due to the reduction of lead C. Side reactions due to the oxidation of chloride D. Side reactions due to the reduction of chloride

B; Salt bridges are ideally composed of inert ions which serve to balance charge in a galvanic cell (or two-chamber electrolytic cell). If a salt bridge is composed instead of lead chloride, the lead cation may take part in side reactions. The cation is most likely to accept electrons (be reduced - choice B is correct) and form lead solid. Chloride is a relatively inert anion and is less likely to participate in side-reactions or form a solid (eliminate choices C and D).

Given the following reaction occurring in a galvanic cell, what is the mass of hydroxide produced when the cell is run for 20 minutes at 3 A? (Faraday's constant = 96,500 C/mol e-) ClO4-(aq) + H2O(l) + Sn2+(aq) → ClO3-(aq) + 2 OH-(aq) + Sn4+(aq) A. 0.34 grams B. 0.63 grams C. 0.17 grams D. 0.41 grams

B; The mass of hydroxide produced when the cell is run for 20 minutes at 3 A is 0.63 grams. Using unit analysis, we can convert from time the current flows into mass of hydroxide as follows: The molar ratio for electrons can most easily be determined by looking at the Sn half-reaction: Sn2+ → Sn4+ + 2 e- Since 2 mol e- are needed to oxidize 1 mol Sn2+ to 1 mol Sn4+, 2 mol e- are transferred in the overall balanced redox reaction as well. Therefore, there is a 2 mol OH- to 2 mol e- ratio.

For the redox reaction 3 MnO2 + 2 Al —> 2 Al2O3 + 3 Mn Which of the following shows the oxidation half-reaction? A. Mn4+ + 4e- —> Mn B. Mn2+ + 2e- —> Mn C. Al —> Al3+ + 3e- D. Al 4+ —> Al6+ + 6e-

C

In an oxidation-reduction, the oxidation number of an aluminum atom changes from 0 to +3. The aluminum atom has been: A. Reduced, and is a reducing agent B. Reduced, and is an oxidizing agent C. Oxidized, and is a reducing agent D. Oxidized, and is a oxidizing agent

C

Rechargeable batteries such as NiCad or lead-acid batteries can operate as both galvanic cells during the discharge cycle, or as electrolytic cells during the recharging cycle. Which one of the following statements accurately compares and contrasts these two cycles? A. Reduction takes place at the positive electrode during recharging, and oxidation takes place at the negative electrode during discharging. B. Reduction takes place at the anode during recharging, but reduction takes place at the cathode during discharging. C. Oxidation takes place at the positive electrode during recharging, but oxidation takes place at the negative electrode during discharging. D. Oxidation takes place at the anode during recharging, but oxidation takes place at the cathode during discharging.

C

The following two half-reactions occur in a voltaic cell: (1) Cr2O72-(aq) + 14 H+(aq) + 6e- 2 Cr3+(aq) + 7 H2O(l) (2) 6 I-(aq) 3 I2(s) + 6e- A. Reaction 2 is an oxidation and occurs at the cathode. B. Reaction 1 is an oxidation and occurs at the anode. C. Reaction 2 is an oxidation and occurs at the anode. D. Reaction 1 is a reduction and occurs at the anode.

C

The oxidation states of chromium and oxygen in K2Cr2O7 are (respectively): A. +8 and -1. B. +7 and -2. C. +6 and -2. D. +6 and -1.

C

When an element is oxidized, it will: A. lose electrons, and its oxidation state will decrease. B. gain electrons, and its oxidation state will decrease. C. lose electrons, and its oxidation state will increase. D. gain electrons, and its oxidation state will increase.

C

Which of the following best characterizes electron flow in an electrolytic cell? A. Electrons flow from anode to cathode due to the intrinsic potential difference between the electrode materials. B. Electrons flow from cathode to anode due to the intrinsic potential difference between the electrode materials. C. Electrons flow from anode to cathode due to the driving force provided by an external source. D. Electrons flow from cathode to anode due to the electrostatic attraction of the respective poles.

C

Which of the following best describes a concentration cell? A. It is an electrolytic cell where current flows once the system has reached equilibrium. B. It is an electrolytic cell with identical reactants and products. C. It is a galvanic cell with two identical electrodes. D. It is a galvanic cell where current flows once the system has reached equilibrium.

C

Using the following half-reaction potentials, Br2(l) + 2e- 2 Br-(aq) E° = 1.07 V F2(g) + 2e- 2 F-(aq) E° = 2.87 V determine the cell potential and whether the following reaction as written is spontaneous under standard conditions: 2 Br-(aq) + F2(g) Br2(l) + 2 F-(aq) Question 17 Answer Choices A. E° = -1.80 V, spontaneous B. E° = +1.80 V, nonspontaneous C. E° = +1.80 V, spontaneous D. E° = -1.80 V, nonspontaneous

C;

Which of the following best expresses the flow of electrons in a voltaic cell? A. Electrons nonspontaneously flow from the anode up their concentration gradient. B. Electrons nonspontaneously flow from the cathode down their concentration gradient. C. Electrons spontaneously flow toward the cathode due to potential difference. D. Electrons spontaneously flow toward the anode due to potential difference.

C;

Peroxide concentration in a sample is often determined by redox titration with I−. The following equation describes the relevant reaction: 2 H+ + H2O2 + 2 I− → I2 + 2 H2O What was the initial concentration of peroxide in a 100 mL sample if 4.2 mL of a 1 M solution of NaI was required to reach the endpoint? A. 0.0042 M B. 0.042 M C. 0.021 M D. 0.0021 M

C; 4.2 mL of a 1 M solution of NaI contains 0.0042 moles of I−. Since two moles of I− are required for each peroxide molecule, there must have initially been 0.0021 mol of peroxide (eliminate choices A and B). In 100 mL this amounts to a concentration of 0.021 M (eliminate choice D).

Each of the following are true of redox titrations EXCEPT: A. the reaction occurring during the titration involves the transfer of electrons between the titrant and analyte. B. at the half-equivalence point, the concentrations of the oxidized and reduced form of the analyte are equal. C. at the equivalence point, the number of moles of titrant added equals the number of moles of analyte at the start of the titration. D. at the equivalence point, the reaction between the analyte and titrant is complete.

C; At the equivalence point, the number of moles of titrant added equals the number of moles of analyte at the start of the titration is not always a true statement about a redox titration. By definition, at the half-equivalence point one-half of the original number of moles of analyte has reacted. For a redox titration that means half the number of moles has either been oxidized or reduced, depending on the titrant used. This means the oxidized form and the reduced form will both have the equivalent of half the original number of moles than at the start. Since both forms are in the same vessel, concentrations will also be equal. A redox titration is a method which uses redox reactions to analyze for an unknown concentration. All redox reactions involve electron transfer. Once the equivalence point is reached, the reaction between analyte and titrant is complete. Addition of more titrant will simply increase the concentration of the titrant without subsequent change to the solution. The final statement is only true if for each half-reaction, the same number of electrons is lost in the oxidation half-reaction as is gained in the reduction half-reaction. When this is not the case, the number of moles of titrant and analyte will not be equal, and they will have different coefficients in the overall balanced redox equation.

In which of the following redox reactions are exactly six moles of electrons transferred? A. 5 SnCl2 + 2 KMnO4 + 16 HCl → 2 MnCl2 + 5 SnCl4 + 8 H2O + 2 KCl B. 2 FeCl3 + Ti(s) → TiCl2 + 2 FeCl2 C. 3 Pd(s) + Na2Cr2O7 + 14 HNO3 → 3 Pd(NO3)2 + 2 Cr(NO3)3 + 7 H2O + 2 NaNO3 D. 2 H2SO3 + H2O2 → H2S2O6 + 2 H2O

C; For any redox reaction, the easiest way to determine the total number of moles of transferred electrons is to determine the change in oxidation states of a given element over the course of the reaction, then multiply by the total number of moles of the element required in the reaction. Each reaction will have two such substances (one that gains electrons and one that loses them), and either substance will work. Choice A shows 5 mol Sn changing from a +2 to a +4 oxidation state, and 2 mol Mn7+ changing to Mn2+, both 10 mol e-transfers (eliminate choice A). Choice B shows 2 mol Fe going from a +3 to a +2 oxidation state, and 1 mole of Ti0 going to Ti2+, both 2 mol e- transfers (eliminate choice B). Choice C shows 3 mol Pb0 going to Pb2+, and 2 mol Cr6+ going to Cr3+, both 6 mol e- transfers. Choice D shows 2 mol S change from a +4 to a +5 oxidation state, while 2 mol O changes from -1 to -2 oxidation states, both 2 mol e- transfers (eliminate choice D).

A car part manufacturer wants to plate each muffler in her warehouse with 6.5 g of chrome (chromium) using an immersion technique with chromium solid and Cr(NO3)3. Given a constant current of 10 A, how long should the manufacturer immerse her mufflers to ensure the appropriate amount of chrome plating? Note: F = 96,500 C/mol e- A. 10 minutes B. 30 minutes C. 1 hour D. 3 hours

C; Given I = C/t, we can rearrange to get t = C/I where I is current (in amps), C is a measure of charge (in coulombs), and t is time (in seconds).

A chemist attempts to assemble a galvanic cell but only has solid silver and silver chloride as reagents. He assembles a cell consisting of two idential beakers each containing a silver electrode and an equimolar solution of silver chloride, but observes no reaction. Which of the following would result in current flow? I. Increasing the temperature of one half-cell II. Adding pure water to one half-cell III. Adding equal quantities of pure water to both half-cells A. I only B. II only C. I and II only D. I, II, and III

C; In order for current to flow, the electrical potential of the two half-cells must differ. Both changing the concentration of the reagents and changing the temperature (see the Nernst equation below) will result in changes in electrical potential and current flow (Roman numerals I and II are true, so choices A and B can be eliminated). However, if both half-cells were diluted equally, the electrical potentials would remain identical and no current flow would occur (Roman numeral III is false, eliminating choice D).

Which of the following half-reactions results in the greatest release of energy? A. Ca2+ + 2 e− → Ca(s) E° = -2.87 V B. Fe2+ + 2 e− → Fe(s) E° = -0.44 V C. Cu+ + e− → Cu(s) E° = 0.52 V D. NO3− + 4 H+ + 3 e− → NO(g) + 2 H2O E° = 0.96 V

D

A battery stops functioning shortly after its initial use. What observation could explain the sudden failure of a battery? A. Increased solid formation at the cathode B. Etching visible on the anode C. Fracture of the anode/cathode barrier D. A slight color change in one of the chambers

C; Mixing of the anode and cathode chambers in a galvanic cell would result in rapid battery death. When the reactants are separated, electrons are forced to follow a specified path to provide useful electrical energy. Breaking this barrier means electrons quickly transfer from reducing agent to oxidizing agent directly, making choice C the best answer. Solid formation at the cathode, etching visible on the anode, and color changes may be part of normal battery function, so we cannot be certain they resulted in battery death (eliminate choices A, B and D).

Given the Nernst equation below, solve for Keq for the oxidation-reduction reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) where E˚ = 1.10 V. E = E (degree) - .059V / n (log Q) A. 1 × 10^‒37 B. 1 × 10^19 C. 1 × 10^37 D. Insufficient information provided

C; The Keq for the oxidation-reduction reaction Zn(s) + Cu2+(aq) ? Zn2+(aq) + Cu(s) where E° = 1.10 V is 1 × 1037. At equilibrium, the electrical potential for a cell is equal to zero and Q = Keq, therefore we can rearrange the Nernst equation as follows:

The chemical reaction occurring in a standard AA battery is as follows: Zn(s) + 2 MnO2(s) + 2 NH4Cl(aq) → Mn2O3(s) + Zn(NH3)2Cl2(aq) + H2O(l) Which of the following represents the (-) pole on the battery? A. MnO2(s) B. Mn2O3(s) C. Zn(s) D. NH4+(aq)

C; The anode in a galvanic cell possesses a negative charge. Ions in solution cannot serve as electrodes, so eliminate choice D. In the above reaction, zinc is being oxidized while manganese is being reduced. Zn(s) is therefore the anode (an ox) and the negative pole on the battery (choice C is correct). Note that determining the new oxidation state of zinc may appear difficult. However, we do not need to know an exact number, only realize that zinc now has a positive oxidation state.

Given the half-reaction reduction potentials in the table below, which of the following reactions represents the overall equation for a lead-acid battery during its recharging cycle? Standard Reduction Half-Reactions Eo (V) Pb2+ + 2e- → Pb -0.13 PbSO4 + 2e- → Pb + SO42- -0.36 PbO2 + 2e- + 4 H+ + SO42- → PbSO4 + 2 H2O +1.69 PbO2 + 2e- + 2 H+ → Pb(OH)2 +1.48 A. PbO2(s) + Pb(s) + 2 H2SO4(aq) → 2 PbSO4(aq) + 2 H2O B. 2 Pb(OH)2(s) → PbO2(s) + Pb(s) + 2 H2O C. 2 PbSO4(aq) + 2 H2O → PbO2(s) + Pb(s) + 2 H2SO4(aq) D. PbO2(s) + Pb(s) + 2 H2O → 2 Pb(OH)2(s)

C; The lead-acid battery operates under acidic conditions not basic ones, so choices that contain Pb(OH)2 can be eliminated. During the recharging cycle, the battery acts as an electrolytic cell, not a voltaic cell. This means that the overall cell potential should be negative, not positive. The two half-reactions which when added together will contain both PbO2 and PbSO4 are the second and third equations in the table. To make these add up to a negative cell potential, the third equation, which has the largest Eo value, must be reversed. This will make PbSO4 a reactant and PbO2 a product.

How many moles of electrons are transferred per two moles of Cr3+(aq) in the redox reaction shown? 2 Cr3+(aq) + 6 Cl-(aq) → 2 Cr(s) + 3 Cl2(g) A. 2 moles B. 3 moles C. 6 moles D. 12 moles

C; To find the number of moles of electrons transferred in the reaction, simply break the overall reaction into each half-reaction and insert the appropriate number of electrons. After appropriate multiplication, the oxidation half-reaction consists of 6 Cl-(aq) → 3 Cl2(g) + 6 e- and the reduction half reaction consists of 2 Cr3+(aq) + 6 e- → 2 Cr(s) (choice C is correct).

Name a biological enzymes that utilizes a disproportionation mechanism

Catalase

What is the oxidation half reaction of a nickel cadmium battery?

Cd (s) + 2 OH- (aq) --> Cd(OH)2 (s) + 2e-

What is electromotive force?

Corresponds to the voltage or electrical potential difference of the cell

The reduction potentials of the reactions below are +1.2 V and -0.8 V, respectively. MnO2(s) + 4 H+ + 2e- → Mn2+(aq) + 2 H2O(l) Zn2+(aq) + 2e- → Zn(s) What is the voltage of a MnO2/Zn(s) cell? A. +0.4 V B. +0.8 V C. +1.6 V D. +2.0 V

D;

A voltaic cell is run for one hour at 2 A and 1.4 grams of solid iron are formed. Assuming a one-to-one mole ratio of solid iron to its oxidized form, iron in which oxidation state served as the reactant? (Faraday's constant = 96,500 C/mol e-) A. Fe0 B. Fe+ C. Fe2+ D. Fe3+

D; A voltaic cell is run for one hour at 2 A and 1.4 grams of solid iron are formed. Assuming a one-to-one mole ratio of solid iron to its oxidized form, iron in its Fe3+ oxidation state served as the reactant. To determine the oxidation state of the reactant iron, we must find the number of moles of electrons transferred per mole of iron. This indicates that the iron began in the +3 oxidation state before accepting three equivalents of electrons to become solid iron.

In the galvanic cell Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s), the addition of excess Zn2+ to the solution at the Zn electrode will have what impact on the voltage? A. Insufficient information provided B. No change C. Increase D. Decrease

D; In the galvanic cell Zn(s)|Zn2+(aq)||H+(aq)|H2(g)|Pt(s), addition of Zn2+ will decrease voltage. The electrochemical cell here gives this overall reaction: Zn(s) + 2 H+(aq) → Zn2+(aq) + H2(g) According to the Nernst equation, as Q increases (as with addition of a product) the voltage decreases. Alternatively, we can relate this to Le Chatelier's principle. As we add more product, the reaction proceeds in reverse to a greater degree. Therefore the forward reaction becomes less spontaneous (voltage decreases) as a product is added.

For double-displacement reactions that do not form a solid salt, there is no ________ ionic equation

net

Fully charged lead-acid batteries utilize metallic Pb0 as the anode, and PbO2as the cathode. Recharging spent batteries using very high voltages has been observed to cause the electrolysis of water into H2 and O2 at the electrodes. Which of the following is true? A. O2 and H2 evolution are observed on the surface of Pb. B. O2 and H2 evolution are observed on the surface of PbO2. C. O2 is evolved on the surface of Pb, and H2 is evolved on the surface of PbO2. D. O2 is evolved on the surface of PbO2, and H2 is evolved on the surface of Pb.

D; During recharge, the formation of Pb0 will be taking place at the cathode, and PbO2 at the anode. Excessively high voltages suggest an excess of free electrons at the cathode, and a strong driving force for the removal of electrons at the anode. The production of H2 from water (2 H2O + 2e- → H2+ 2 OH-) is a reduction reaction, and will require these highly energetic electrons. Therefore it will form on the surface of Pb (eliminate choices B and C). The conversion of water to O2 (2 H2O → O2 + 4e- + 4 H+) is an oxidation, requiring a large oxidative force, such as present at the surface of PbO2 at excessively high recharge voltages (eliminate choice A).

For the reaction below, which of the following statements is true? 2 Au + 3 Fe2+ —> 2 Au3+ + 3Fe Au3+ + 3e- --> Au (E = 1.50) Fe2+ + 2e- --> Fe (-.44) A. The rxn is spon, because its cell voltage is (+) B. The rxn is spon, because its cell voltage is (-) C. The rxn is nonspon, because its cell voltage is (+) D. The rxn is nonspon, because its cell voltage is (-)

D; Eliminate B and C; (+) voltage is spon and (-) cell voltage is nonspon. The half reactions are 2(Au —> Au3+ + 3e-) E = -1.50V 3(Fe2+ + 2e- —> Fe) E = -.44V Iron is reduced Silver is oxidized

Given the standard reduction potentials below, if a galvanic electrochemical cell were prepared using these two half-reactions, what would the values of E°cell and ΔG° be, respectively? [AuCl4]- + 3 e− → Au(s) + 4 Cl- E° = 0.930 VBr2(aq) + 2 e− → 2 Br- E° = 1.087 V A. 2.017 V and 1,170 kJ B. 1.401 V and -45.5 kJ C. -0.157 V and 45.5 kJ D. 0.157 V and -90.9 kJ

D; Galvanic cells are spontaneous electrochemical cells, so the E°cell must be positive (eliminate choice C); ΔG° must be negative for a spontaneous process (eliminate choice A). To create a positive standard cell voltage, the gold half-reaction should be reversed and become an oxidation half-reaction. This will make the E°cell = +1.087 + (-0.930 V) = +0.157 V (eliminate choice A, choice D is correct). Note that to calculate ΔG°, apply the equation ΔG° = -nFE°cell. The total number of electrons transferred, n, will be 6 based on the balanced equation from the half reactions: (2 Au(s) + 8 Cl- + 3 Br2(aq) → 6 Br- + 2 [AuCl4]-). This makes ΔG° = -(6 mol e-)(96,500 C/mol e-)(+0.157 V) ≈ -(6)(100,000)(0.16) ≈ -96,000 J, or just less than -96 kJ via estimation.

A type of rechargeable battery using vanadium species as redox couples cycles between VO2+ and VO2+ at one electrode and between V2+ and V3+ at the other. Hypothetically, if V3+ is produced during discharge, which of the following is consumed during recharge? A. VO2+only B. Both VO2+ and V3+ C. V2+ only D. Both V3+ and VO2+

D; If V3+ is produced during discharge, this means that the V2+/V3+ couple must be the anode in discharge (V2+ → e- + V3+). As such, during recharging this electrode must be the cathode, so the reverse of the process must occur, meaning V2+ will be produced, not consumed (eliminate choice C). This also means the VO2+/VO2+ couple is at the anode during the recharging process, where oxidation occurs. Since vanadium in VO2+ is in the +4 oxidation state and in VO2+ is +5, the anodic reaction would consume VO2+ (2 OH- + VO2+→ VO2+ + e- + H2O) (eliminate choices A and B).

In a fully charged NiCad battery, the anodic and cathodic reactions are as follows: Cd + 2 OH- → Cd(OH)2 + 2e- (anode) 2 NiO(OH) + 2 H2O + 2e- → 2 Ni(OH)2 + 2 OH- (cathode) Which of the following statements is FALSE? A. Ni2+ is oxidized as the battery is recharged. B. Cd2+ is reduced as the battery is recharged. C. Cd is oxidized as the battery is discharged. D. Ni2+ is reduced as the battery is discharged.

D; In discharge Cd is converted from Cd0 to Cd2+, indicating a loss of electrons, or oxidation (eliminate choice C). That means during recharge the reverse process will happen, so Cd2+ is reduced (eliminate choice B). During discharge, the NiO(OH) is converted to Ni(OH)2, which involves a conversion of Ni from the +3 oxidation state to the +2 state, or a reduction (choice D is the false statement). Therefore Ni2+ is oxidized during recharge (eliminate choice A).

In the earth's crust, iron is predominantly found as iron ore (a mixture of iron oxides) while gold is found in its neutral, metallic form. Which of the following statements is most likely true? A. Iron oxides are composed of neutral iron bound to oxygen. B. Gold is more easily oxidized than iron in the primarily aerobic environment of Earth. C. The reduction potential of metallic gold is larger than that of cationic iron. D. The reduction potential of cationic gold is larger than that of cationic iron.

D; In the earth's crust, iron is predominantly found as iron ore (a mixture of iron oxides) while gold is found in its neutral, metallic form. It is most likely true that the reduction potential of cationic gold is larger than that of cationic iron. The answer choice, "Gold is more easily oxidized than iron in the primarily aerobic environment of Earth" may be eliminated because it is the opposite of what is stated in the question (i.e., iron is typically oxidized while gold is not). Taking neutral gold to a -1 oxidation state would require significantly more energy compared to reducing cationic iron. Iron oxides are the result of iron metal being oxidized to a cationic form by O2 from the atmosphere. If iron is oxidized and gold is not, gold must have a more negative oxidation potential, and hence cationic gold must have a more positive reduction potential than cationic iron.

In the galvanic cell Ni|Ni2+||Fe3+, Fe2+|Pt, what best describes the direction of current flow and flow of Na+ from the salt bridge? A. Current flows from Ni to Pt, and Na+ flows toward Pt. B. Current flows from Ni to Pt, and Na+ flows toward Ni. C. Current flows from Pt to Ni, and Na+ flows toward Ni. D. Current flows from Pt to Ni, and Na+ flows toward Pt.

D; In the galvanic cell Ni|Ni2+||Fe3+, Fe2+|Pt, the current flows from Pt to Ni and Na+ flows toward Pt. In electrochemical cells, electrons flow from anode to cathode (therefore, due to the sign convention, current flows from cathode to anode). In this cell, Ni2+ is being oxidized and nickel is the anode, while Fe3+ is being reduced at a platinum cathode. Thus current flows from the Pt cathode to the Ni anode, allowing us to eliminate two answer choices. Na+ meanwhile travels from the salt bridge to the cathode to offset the charge imbalance generated by the current (cations migrate to the cathode).

A potassium chloride salt bridge connects the two halves of an electrochemical cell. Which of the following best describes the path the salt bridge ions follow? A. Potassium flows to the anode, and chloride flows to the cathode. B. Potassium flows to the anode, and chloride flows to the anode. C. Potassium flows to the cathode, and chloride flows to the cathode. D. Potassium flows to the cathode, and chloride flows to the anode.

D; Salt bridges are responsible for neutralizing charge imbalance that results from the reaction occurring in a galvanic cell (or two-chamber electrolytic cell), so the cations and anions must travel to opposite electrodes. This eliminates choices B and C. Given that electrons flow from the anode to the cathode, anions from the salt bridge must travel to the anode and cations from the salt bridge must travel to the cathode to balance the change in charge (choice D is correct).

Which of the following would result in a change in the electrical potential of a galvanic cell consisting of solid silver and copper electrodes in solutions of their respective ions? A. Increasing the amount of solid silver anode in a reaction chamber B. Changing the identity of the salt in the salt bridge C. Changing the wire to another conductive metal D. Increasing the temperature of the cell

D; The Nernst equation, which expresses the electrical potential of a cell, can be written as: Thus, changing the temperature will affect E for the cell in question (and also Keq and ΔG - choice D is correct). Similarly, changes in reagent concentrations (affecting Q) would also affect the electrical potential. The remaining factors are not components of the Nernst equation and will not affect electrical potential. Specifically, changing the amount of solid silver anode (this is not part of the Q expression as it is a solid), the identity of the salt bridge, or the wire connecting the anode/cathode would have limited impact on potential.

Which of the following cells generates the greatest amount of energy per Coulomb? Zn2+ + 2e- → Zn E˚ = ‒0.76 Al3+ + 3e- → Al E˚ = ‒1.66 Ag+ + e- → Ag E˚ = +0.80 Cu+ + e- → Cu E˚ = +0.52 A. Al|Al3+||Zn2+|Zn B. Cu|Cu+||Ag+|Ag C. Ag|Ag+||Cu+|Cu D. Zn|Zn2+||Cu+|Cu

D; The cell that generates the greatest amount of energy per Coulomb is Zn|Zn2+||Cu+|Cu. Energy per Coulomb, or J/C, is another form of the unit for the volt, thus this question is asking us for the galvanic cell with the greatest potential difference. To determine the potential difference of each listed cell, we must sum the two half reactions comprising the cell. The cell with the greatest potential is Zn|Zn2+||Cu+|Cu with E°= 0.76 V + 0.52 V = 1.28 V.

A chemist creates a concentration cell with equal volume half-cells (Cu|Cu+ (0.030 M)||Cu+ (1.0 M)|Cu), allows it to react for several minutes, and analyzes the solutions before the system has reached equilibrium. What best describes the concentrations seen in the cell at this time? A. Anode = 0.030 M Cu+; cathode = 1.0 M Cu+ B. Anode = 1.0 M Cu+; cathode = 0.030 M Cu+ C. Anode = 0.020 M Cu+; cathode = 1.01 M Cu+ D. Anode = 0.100 M Cu+; cathode = 0.93 M Cu+

D; The final concentrations seen in the cell will be 0.100 M Cu+ at the anode and 0.93 M Cu+ at the cathode. Concentration cells are galvanic cells with a potential difference between the half-cells due to differing concentrations. The anode is the site of oxidation, thus we can expect the concentration of Cu+ to increase. Conversely, the concentration of Cu+ should decrease at the cathode, the site of reduction. The reaction proceeds until both half-reactions have the same concentration, resulting in a voltage difference of zero. Therefore the only answer choice that would be possible is 0.100 MCu+ at the anode and 0.93 M Cu+ at the cathode. Note that 1.0 M Cu+ at the anode and 0.030 M Cu+at the cathode would not be possible as the concentration at the anode would not continue to increase (or the concentration at the anode decrease) once the voltage has fallen to zero.

The oxidation states of sulfur in SO42- and SO32- are (respectively): A. +8 and +6. B. +8 and +4. C. +6 and +6. D. +6 and +4.

D; The oxidation states of sulfur in sulfate and sulfite can be determined using the typical oxidation state rules. Oxygen maintains an oxidation state of -2, totaling -8 and -6, respectively, while the overall anions have a charge of -2. This means that sulfur must have charges of +6 and +4, respectively, to cancel out the negative charge and result in an overall -2 for the anion, making choice D correct.

Which of the following best describes the reduction half-reaction in the following reaction? Cu(s) + 2 NO3-(aq) + 4 H+(aq) → Cu2+(aq) + 2 NO2(g) + 2 H2O(l) A. Cu(s) + NO3-(aq) → Cu2+(aq) + NO2(g) B. Cu(s) → Cu2+(aq) + 2 e- C. 4 H+(aq) + O2(g) + 4 e- → 2 H2O(l) D. 2 NO3-(aq) + 2 e- + 4 H+(aq) → 2 NO2(g) +2 H2O(l)

D; This question is asking for a reduction half-reaction which therefore must involve only accepting electrons (choices A and B are wrong). The reaction described here involves the reduction of nitrogen in nitrate and the oxidation of copper metal to copper(II) cation. While answer choice C describes the oxidation of oxygen, this reaction is not a part of the overall reaction as oxygen gas is not involved (choice C is wrong). The reduction of nitrate to nitrogen dioxide gas describes the reduction half-reaction in the overall reaction (choice D is correct).

The final products of the electrolysis of aqueous NaCl are most likely: A. Na(s) and Cl2(g) B. HOCL(aq) and Na(s) C. Na(s) and O2(g) D. NaOH(aq) and HOCl(aq)

D; Use POE: Choices A, B, and C list metallic sodium as a final product. That is a problem because we are in an aqueous medium, we know that sodium metal reacts violently in water to form NaOH and hydrogen gas and fire.

Delta G, at non-standard conditions

Delta G = delta G degree + RT lnQ

Nerst Equation

E = E(degree) - (RT/nF) ln Q

How do you find changes in cell potential?

E(degree) = E(cathode) - E (anode)

Reducing agents are also called

Electron donors

If a 2 A is applied for 2 min, how many moles of Cu2+ can be plated out as Cu metal (F = 96,485 C/mol)?

I = nF / t n = It / F n = (2A)(120 seconds) / 96,500 C n = about .0025 moles e-

Given current and time, what is the equation to find the number of electrons?

I = nF / t I = current n = moles of e- F = Faraday's constant t = time (in seconds)

Identify the oxidation states of the relevant atoms, the oxidizing agent, and the reducing agent. 2 KI + H2 —> 2 K + 2 HI

KI: K(+1), I(-1) H2 (0) 2K (0) 2 HI: H(1), I(-1) Oxidizing agent: K Reducing agent: H2

What is the conductive material in a nickel cadmium battery?

KOH

How do redox titrations work?

Monitor redox reaction instead of protons

What is the cathode of a nickel cadmium battery?

nickel III oxide-hydroxide

Mnemonic to remember oxidation and reduction

OIL RIG Oxidation is losing (electrons) Reduction is gaining (electrons)

What is a decomposition reaction?

One product breaks down into two or more species

What is the anode in a lead acid battery

Pb

What is the oxidation hold reaction at the lead anode?

Pb(s) + HSO4-(aq) --> PbSO4 (s) + H+ + 2e-

What is the overall net equation of a lead acid battery?

Pb(s) + PBO2 (s) + 2 H2SO4 (Aq) --> 2 PbSO4 (s) + 2 H2O

What is the cathode in a lead acid battery

PbO2

What is the reduction half reaction at the lead (IV) oxide cathode

PbO2 (s) + SO4 2- (aq) + 4 H+ + 2e- --> PbSO4 (s) + 2 H2O

If 5 amps of current flowed in NaCL electrolytic cell for 1930 seconds, how much sodium metal and chlorine gas would be produced?

Q = it Q = 5 amps x 1930 s Q = 9650 coulombs 9650 C x 1 mole/96500 C = .1 mole e- Na+ e --> Na. .1 mole of electrons would give 1 mole Na, therefore .1 mol of electrons gives .1 mole of Na. Since the molar mass of sodium is 23 g/mol, we get 2.3 g of sodium 2 Cl- --> Cl2 + 2e-, for every 1 mole of electrons lost, we get .5 mole of Cl2. Since step 2 told us that .1 mol of electrons were liberated at the anode, .05 mol of Cl2(g) was produced. Because the molar mass of Cl2 is 71 g/mol, we'd get 3.55 g of chlorine gas.

What is a nickel-cadmium battery?

Rechargeable batteries

What is the oxidation number of an atom?

The oxidation number of an atom is the charge that would exist on an individual atom if the bonding were completely ionic.

What is electroplating?

The process of depositing a thin layer of metal on an object during electrolysis.

What is a Daniell cell?

The specific galvanic cell with zinc and copper bound to sulfate. Both zinc and copper ions are +2

What is electroplating?

The use of electrolysis to put a thin layer of metal on the surface of an object

What are lead-acid batteries?

Type of rechargeable battery

Reduction potential is measured in

Volts

What is the cell diagram for Daniell cell?

Zn (s) | Zn2+ (1M) || Cu2+ (1M) | Cu (s)

What is an oxidation-reduction reaction?

a chemical reaction in which electrons are exchanged between atoms

What does a single line in a cell diagram signify?

a phase boundary

What does a vertical line in a cell diagram signify?

a salt bridge

What connects the two aqueous solutions in a galvanic cell?

a slat bridge (such as KNO3)

What is a Faraday?

amount of charge contained in one mole of electrons, approximately 96,500 coulombs

An electrolytic cell uses what?

an external voltage (such as a battery) to create an electric current that forces a nonspontaneous redox reaction to occur.

For all electrochemical cells, the electrode where oxidation occurs is the

anode

For galvanic and electrolytic cells, anions migrate to the

anode

For galvanic and electrolytic cells, oxidation occurs at the

anode

How do electrons flow in a galvanic cell?

anode to cathode

For galvanic and electrolytic cells, cations migrate to the

cathode

For galvanic and electrolytic cells, reduction occurs at the

cathode

The electrode where all reductions take place is the

cathode

When the electrodes are connected by a wire, zinc atoms in the anode are oxidized and the electrons travel through the wire to the

cathode

What does a positive emf signify?

cell is able to release energy (spontaneous)

What does a negative end signify?

cell must absorb energy (nonspontaneous)

What is the conductive material in a lead acid battery?

concentrated H2SO4

What are electrochemical cells?

contained systems in which oxidation-reduction reactions occur

In a concentration, the electrodes are the same but the half cells have different what?

different ion concentrations

What charging, a lead acid battery is a ________- circuit

electrolytic

Molten NaCl discharging galvanic or electrolytic? Anode material: Anode charge: Cathode material: Cathode charge:

electrolytic any (+) any (-)

Ni-Cd, charging galvanic or electrolytic? Anode material: Anode charge: Cathode material: Cathode charge:

electrolytic CdOH2 (+) NiOH2 (-)

Lead acid charging galvanic or electrolytic? Anode material: Anode charge: Cathode material: Cathode charge:

electrolytic PbSO4 (+) PbSO4 (-)

Oxidizing agents are also called

electron acceptors

For galvanic and electrolytic cells, electrons flow

from anode to cathode

What is reduction?

gain of electrons, loss of oxygen, gain of hydrogen

Ni-Cd, discharging galvanic or electrolytic? Anode material: Anode charge: Cathode material: Cathode charge:

galvanic Cd (-) NiO(OH)2 (+)

Lead acid discharging galvanic or electrolytic? Anode material: Anode charge: Cathode material: Cathode charge:

galvanic Pb (-) PbO2 (+)

Daniell cell discharging galvanic or electrolytic? Anode material: Anode charge: Cathode material: Cathode charge:

galvanic Zn (-) Cu (+)

What are the three types of electrochemical cells?

galvanic (spontaneous) electrolytic (nonspontaneous) concentration (spontaneous)

all non rechargeable batteries are

galvanic cells

In a galvanic cell, two electrodes of distinct chemical identity are placed in separate compartments, which are called

half cells

What are double displacement reactions?

involves switching two counter ions

What are spectator ions?

ions that do not participate in a reaction

Nerst equation at standard conditions reduces to what?

log K = nE / .0592

What is oxidation?

loss of electrons, gain of oxygen, loss of hydrogen

Double displacement reactions are also called

metathesis reactions

In electrolytic cells, the cathode is ________

negative

Reducing agents have the most _________ reduction potential

negative

The anode in a galvanic cells is

negative

The stronger reducing agents have a more _________ reduction potential

negative