Electrochemistry

The oxidation number of Mn in Mn3(PO4)2 is

+2 It is important to recognize the subscript changes the amount of elements in the compound. Using the known rules, we know that O is -2 and Mn, since it is in group 2 would be +2. Mn= 2 x 3= +6 and O= -2 x 4 x 2= -16 P= 2 x 5= 10

What is the potential of this cell? Enter your answer to the hundredths place (x.xx V). Pt | Pb2+ (0.10 M), Pb4+ (0.10 M) || Fe2+ (0.025 M) | Fe (activity series)

-2.16 The shorthand notation always lists the anode on the left and the cathode on the right, so this tells you that the Pb2+ is being oxidized and the Fe2+ is being reduced. Pt | Pb2+ (0.10 M), Pb4+ (0.10 M) || Fe2+ (0.025 M) | Fe The Nernst equation is Ecell = Eocell - 0.05916/n logQ, so you need to determine Eocell, n, and Q in order to calculate the cell potential of this cell. Eocell = Eored - Eoox Q = [products]/[reactants] n= the number of electrons transferred in the balanced equation.

What is the standard free energy (ΔG°) of the reaction represented by the cell shorthand notation Pt | Fe2+, Fe3+ || Ag+ | Ag?

-2.89 Iron is the anode and the oxidation reaction is Fe^2+ → Fe^3+ + e- Silver is the cathode and the reduction reaction is Ag+ + e- → Ag Determine how many electrons are transferred in the balanced reaction and calculate E°cell = E°red - E°ox. Then use ΔG° = -nFE°cell to calculate ΔG°.

For this reaction, which is the correct unbalanced oxidation half reaction? Al (s) + Ni^2+ (aq) ⟶Ni (s) + Al^3+ (aq) A. Al (s) ⟶ Al^3+ (aq) B. Ni^2+ (aq) ⟶ Al^3+ (aq) C. Ni (s) ⟶ Ni^2+ (aq) D. Ni^2+ ⟶ (aq) Ni (s) E. Al (s) ⟶ Ni (s) F. Al^3+ (aq) ⟶ Al (s)

A. Al (s) ⟶ Al3+ (aq) 1. Assign oxidation numbers to all the elements in the reaction. Al (s) + Ni2+ (aq) ⟶ Ni (s) + Al3+ (aq) 0 2+ 0 3+ 2. Aluminum is being oxidized so that will be the oxidation half reaction. Al (s) ⟶ Al^3+ (aq)

Balance this equation in basic solution. What is the coefficient of nitrate ion? MnO4− (aq) + NO2− (aq) ⟶ MnO2 (s) + NO3− (aq) A. 1 B. 3 C. 6 D. 9 E. 4

B. 3 1. Separate the overall equation into 2 half-reactions. MnO4− (aq) + NO2− (aq) ⟶ MnO2 (s) + NO3− (aq) Oxidation half reaction = Reduction half reaction= 2. Balance each half reaction by balancing the charge using electrons. For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding a H+ ion to the side deficient in positive charge. 3. Balance the equation.

Which species will REDUCE Ag+ but not Fe2+? A. Co^2+ B. H2 C. K D. Cr

B. H2 Ag+ + e- ⟶ Ag E° = +0.80 V Fe2+ + 2e- ⟶ Fe E° = -0.44 V For these species to reduce, they must act as the cathode. Therefore, all of the answer choices must act as an anode. Remember... E° = E cathode - E anode H2 has an E° of 0.00, and gives a positive E°cell when combined with Ag+ as the cathode but a negative E°cell when combined with Fe2+ as the cathode.

In a galvanic cell... A. oxidation takes place at the cathode B. oxidation and reduction take place at the same time, but at different electrodes C. electrolytes are added to carry electrons between electrodes D. electrical energy is used to reverse spontaneous chemical reactions

B. oxidation and reduction take place at the same time, but at different electrodes The reaction in a galvanic cell is spontaneous and does not need to be driven by a battery. Both reduction and oxidation occur simultaneously, oxidation at the anode and reduction at the cathode.

If I told you a redox reaction was non-spontaneous, what could you tell me about the free energy and equilibrium constant of the reaction and about an electrochemical cell that operated using that reaction?

E° < 0 ΔG° > 0 K < 1

True or False: Solid nickel will be easily oxidized in a solution of aqueous sodium sulfate according to an activity series.

False Sodium is higher on the activity series than nickel is, so the electrons in a sodium atom have a higher free energy than those in a nickel atom. Therefore, the electrons in nickel will stay where they are.

Based on the following reaction, what species is the oxidizing agent? How many electrons were transferred as it is balanced? 2Fe^2+ (aq) + H2O2 (aq) → 2Fe^3+ (aq) + 2OH- (aq)

H2O2, 2

Consider the voltaic cell: Pt | Sn2+ (0.10 M), Sn4+ (0.0010 M) || Ag+ (0.010 M) | Ag Sn4+ + 2e- ⟶ Sn2+ E° = +0.15 V Ag+ + 1e- ⟶ Ag(s) E° = +0.80 V The electrons flow in the external circuit from...

Pt to Ag In cell notation, the anode reaction appears on the left side of the salt bridge and the cathode on the right side. Electrons flow between the electrodes specifically. As this is a voltaic cell, electrons flow from the anode via the Pt electrode through the external circuit to the Ag cathode.

In the reaction of thiosulfate ion with chlorine gas in an acidic solution, what is the reducing agent? Cl2(g) + S2O3^2-(aq) ⟶ Cl^-(aq) + SO4^2-(aq)

S2O3^2- Remember, the reducing agent is itself oxidized. In S2O3^2-, the oxidation number of sulfur is +2. In SO4^2-, the oxidation number of sulfur is +6. Therefore, S2^+ is being oxidized in this reaction. However, remember that a reducing agent or oxidizing agent includes the entire species that contains the atom being oxidized or reduced, respectively. Therefore, as S2^+ is being oxidized, S2O3^2- is the reducing agent.

The electrode in an electrochemical cell where oxidation occurs is called the ___________ and the electrode where reduction is occurring is called the __________.

anode cathode An ox, red cat

In the activity series the half-reactions are written as oxidations and the lower free energy metals are at the bottom. In a table of standard reduction potentials, the half-reactions are written as reductions, and the metals with lower free energies are:

at the top

Balance the following reaction in basic conditions and answer the following questions: Ca2+ (aq) + C(s) + ClO2 (g) → CaCO3(s) + ClO2- (aq) What is the oxidation state of C in CaCO3(s)? What is the total number of electrons transferred?

+4, 4

Consider the half-reactions: Mn2+ + 2e- ⟶ Mn E° = -1.029 V Ga3+ + 3e- ⟶ Ga E° = -0.560 V Fe2+ + 2e- ⟶ Fe E° = -0.409 V Sn2+ + 2e- ⟶ Sn E° = -0.136 V Which of these redox couples could not be used to make a voltaic cell?

2 Ga^3+ 3 Fe ⟶ 2 Ga + 3 Fe^2+ Only 2Ga3+ + 3Fe ⟶ 2Ga + 3Fe2+ has a negative E°cell, so it is the only reaction here that is non-spontaneous.

A galvanic cell is constructed using a magnesium electrode in a 1.0 M Mg(NO3)2 solution and a silver electrode in a 1.0 M AgNO3 solution. What is the standard potential of the cell at 25°C? A. -3.17 V B. +3.16 V C. +3.97 V D. -1.57 V

B. + 3.16 v Since the reduction of Ag+(aq) to Ag(s) has a higher potential than the reduction of Mg2+(aq) to Mg(s), Ag+(aq) will be reduced at the cathode in a galvanic cell. (You could also determine this from the activity series if reduction potentials were not provided.) Mg(s) will therefore be oxidized at the anode. Ecell = Ered - Eox = 0.80 - (-2.36) = +3.16 v

For this reaction, what is the coefficient of Al (s) in the overall balanced equation? Al (s) + Ni^2+ (aq) ⟶Ni (s) + Al^3+ (aq) A. 1 B. 2 C. 3 D. 4 E. 5

B. 2 1. Separate the overall equation into 2 half-reactions. Al (s) + Ni2+ (aq) ⟶ Ni (s) + Al3+ (aq) 0 2+ 0 3+ Oxidation half reaction = Al (s) ⟶ Al^3+ Reduction half reaction= Ni^2+ (aq) ⟶ Ni (s) 2. Balance each half reaction by balancing the charge using electrons. 2 x (Al (s) ⟶ Al^3+ + 3e-)= 2Al ⟶ 2Al^3+ + 6 e- 3 x (2e- + Ni^2+ (aq) ⟶ Ni (s))= 6e- + 3 Ni^2+ ⟶ 3 Ni (s) 2 Al (s) + 3 Ni^2+ (aq) ⟶ 2 Al^3+ + 3 Ni (s)

What is the E° for the following electrochemical cell where Zn is the cathode? Fe | Fe2+ (1.0 M) || Zn2+ (1.0 M) | Zn E°(Zn) = -0.76 E°(Fe) = -0.44 A. + 0.32 B. -1.20 C. -0.32 D. +1.20

C. -0.32 E°cell = E°cathode - E° anode= -0.76 -(-0.44)= -0.32

Find the standard emf of the given cell diagram: Cu(s) | Cu2+(aq) || Au+(aq) | Au(s) Cu2+ + 2e- ⟶ Cu E° = +0.34 V Au+ + e- ⟶ Au E° = +1.69 V A. -1.35 V B. +2.03 V C. -2.03 V D. +1.35 V

D. +1.35 V E°cell = E°cathode - E° anode= +1.69 -(-0.34)= +1.35 v

If the standard potentials for the couples Cu2+|Cu, Ag+|Ag, and Fe2+|Fe are +0.34, +0.80, and -0.44 V respectively, which is the strongest reducing agent? A. Cu^2+ B. Ag^+ C. Cu D. Ag E. Fe F. Fe^2+

E. Fe The best reducing agents are good oxidizers. Good oxidizers are bad reducers and, therefore, have low E°red. The pair with the lowest E°red is Fe2+|Fe. In this pair, Fe would oxidize to Fe2+. Therefore, Fe is the species that oxidizes and is the best reducing agent.

True or false: Metals at the bottom of the activity series are most easily oxidized.

False, at the top

Is ZnCO3 (s) ⟶ ZnO (s) + CO2 (g) a redox reacrtion?

No, nothing is being oxidized or reduced.

The oxidation number of hydrogen in AlH3 is

-1 AlH3 H= -1 x 3= -3 Al= +3

What would the voltage be of this electrochemical cell under standard conditions? Fe | Fe2+ ║Pb4+, Pb2+ | Pt A. 1.62 v B. 2.43 v C. 2.11 v D. -2.43 v E. -2.11 v

2.11 V

Electrochemical cell

gives an electric current with a steady voltage as a result of an electron transfer reaction made up of two half-cells joined by a wire and a salt bridge

In which of these compounds does phosphorous have an oxidation number of +5? A. PH3 B. CoPO4 C. Mn3(PO4)2 D. PCl3 E. P2O5

B, C, E Why B is Correct- First, remember that polyatomic ions in the compound must have oxidation numbers that equal the overall charge. PO4 has a 3- negative charge so the oxidation numbers of phosphorus and oxygen must equal negative 3. Using known rules, Oxygen has an oxidation number of 2-. Multiply -2 x 4 because there are 4 oxygen's in the ion to get a total charge of -8. Since P + O4 must equal -3, phosphorus must have an oxidation number of =+5. Co= +3 P= +5 O= -2 Why C is correct- It is important to recognize the subscript changes the amount of elements in the compound. Using the known rules, we know that O is -2 and Mn, since it is in group 2 would be +2. Multiplying through with the subscript. we get: Mn= 2 x 3= +6 and O= -2 x 4 x 2= -16 To calculate the oxidation number of Phosphorus, remember that there are 2 so that should be accounted for. +6 + 2x + -16= 0 x must be equal to +5 Why E is correct- O = -2 x 5= -10 P= 2x= 10 to equal 0. X= -5

What is the standard cell potential of a battery made from the half reactions below? 2H+ + 2e- ⟶ H2 E° = 0.00V O2 + 4H+ + 4e- ⟶ 2H2O E° = +1.23 V A. -2.46 B. -1.23 C. 1.23 D. -2.46

C. 1.23 E°cell = E°cathode - E° anode= 1.23-0= 1.23

In an electrolytic cell, the negative terminal is the (cathode/anode) and is the site of the (oxidation/reduction) half-reaction.

Cathode, reduction

Consider the cell: Zn(s) | Zn2+(aq) || Cl-(aq) | AgCl(s) | Ag(s). Calculate E° using the standard potentials list. A. -1.20 v B. + 1.20 v C. + 0.54 v D. + 0.98 v

D. + 0.98 v Zn2+ + 2e- ⟶ Zn E° = -0.76 Ag+ + e- ⟶ Ag E° = +0.22 E°cell = 0.22-(-0.76) = 0.98 V

Which species will oxidize Cr2+ (E°red = -0.407) but not Mn2+ (E°red = +1.224)? A. Pb^4+ (E°red = +1.68) B. Zn^2+ (E°red = -0.762) C. Fe^2+ (E°red = -0.771) D. O3 in acid (E°red = +2.076) E. V^3+ (E°red = -0.255)

E. V^3+ (E°red = -0.255) To make Cr2+ or Mn2+ oxidize, they must act as the anode. Remember: 𝐸∘𝑐𝑒𝑙𝑙=𝐸𝑐𝑎𝑡ℎ𝑜𝑑𝑒−𝐸𝑎𝑛𝑜𝑑𝑒 E°cell= E cathode−E anode Plug in the standard reduction potentials for Cr2+ and Mn2+ for the E anode and the standard reduction potential for each answer as the E cathode. We're looking for an answer choice which will give us a positive E°cell when combined with Cr2+ (meaning Cr2+ will be spontaneously oxidized), but give us a negative E°cell when combined with Mn2+ (meaning Mn2+ will NOT be spontaneously oxidized). The only answer choice that does this is V3+. Zn2+ and Fe2+ cannot oxidize either metal, while Pb4+ and O3 in acid can oxidize both.

Standard reduction potential

The tendency of a species to be reduced, as measured at 25 C when reacting species are of 1M concentration or 1atm partial pressure.

Balance the reaction using oxidation and reduction half-reactions. What is the smallest possible integer coefficient of So4^2- in the combined balanced equation? Cl2(g) + S2O3^2-(aq) ⟶ Cl-(aq) + SO4^2-(aq)

2 1. Write the balanced reduction and oxidation half-reactions. To balance the oxygens and hydrogens, add H+ to one side and H2O to the other to denote an acidic solution. Be sure to account for the number of electrons gained or lost as well. Reduction half-reaction: Cl2 (g) + 2 e- ⟶ 2 Cl- (aq) Oxidation half-reaction: S2O3^2- (aq) + 5 H2O (l) ⟶ 2 SO4^2- (aq) + 10 H+ (aq) + 8e- 2. Multiply the reduction half reaction by 4 to balance electrons between the 2 half reactions. 4 Cl2 (g) + S2O3^3- + 5 H2O (l) ⟶ 8 Cl- (aq) + 2 SO4^2- (aq) + 10 H+ (aq)

Calculate the equilibrium constant for the reaction represented by the cell shorthand notation. Pt | Fe2+, Fe3+ || Ag+ | Ag

3.22 Iron is the anode and the oxidation reaction is Fe2+ → Fe3+ + e- Silver is the cathode and the reduction reaction is Ag+ + e- → Ag The chemical reaction is Fe2+ + Ag+ → Fe3+ + Ag E°cell = E°red - E°ox = 0.80 v - 0.77 v = 0.03 v. One electron is transferred in the overall reaction, so n = 1.

Balance the skeletal equation of hydrazine with chlorate ions, shown below: N2H4(g) + ClO3-(aq) ⟶ NO(g) + Cl-(aq) The reaction takes place in basic solution. What is the smallest possible integer coefficient of ClO3^- in the balanced equation? Identify the reducing agent.

4, N2H4 Write the balanced reduction and oxidation half-reactions. To balance the oxygens and hydrogens, add OH- to one side and H2O to the other to denote a basic solution. Be sure to account for the number of electrons gained or lost as well. Oxidation half-reaction: N2H4(g) + 8OH-(aq) ⟶ 2NO(g) + 6H2O(l) + 8e- Reduction half-reaction: ClO3-(aq) + 3H2O(l) + 6e- ⟶ Cl-(aq) + 6OH-(aq) Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4 to balance the e- between the two half-reactions. Then, add the two together to find the full, balanced reaction: 3N2H4(g) + 4ClO3-(aq) ⟶ 6NO(g) + 6H2O(l) + 4Cl-(aq) N2H4 is the reducing agent since N2H4 is oxidized to NO.

Balance the following reaction in acidic conditions. How many total electrons are transferred in this reaction? What is the oxidation state of V in V2O5? What is the reducing agent? FeO (aq) + V2O5 (aq) → Fe2O3 (aq) + VO(aq)

6, 5, FeO

Consider the cell reaction represented by the skeletal equation: Mn(s) + Ti2+(aq) ⟶ Mn2+(aq) + Ti(s) What is the proper shorthand notation for this reaction? A. Mn(s) | Mn2+(aq) || Ti2+(aq) | Ti(s) B. Ti(s) | Ti2+(aq) || Mn2+(aq) | Mn(s) C. Ti2+(aq) | Ti(s) || Mn(s) | Mn2+(aq) D. Mn2+(aq) | Mn(s) || Ti(s) | Ti2+(aq)

A. Mn(s) | Mn2+(aq) || Ti2+(aq) | Ti(s) Oxidation half-reaction: Mn (s) ⟶ Mn^2+ (aq) + 2e- Reduction half-reaction: Ti^2+ (aq) + 2 e- ⟶ Ti (s) When writing the cell diagram for a reaction, remember to write the oxidation half-reaction on the left and the reduction half-reaction on the right, separated by a double vertical line.

An electrochemical cell is constructed using the following spontaneous redox reaction: X(s) + Y2+(aq) ⇌ X2+(aq) + Y(s) When the Y2+ and X2+ concentrations are equal, the cell generates a positive voltage of 1.00 v. How will the cell potential change if the concentrations are changed so that [X2+] = 10 × [Y2+]? A. The voltage of the cell will decrease. B. The voltage of the cell will increase. C. The voltage of the cell will increase.

A. The voltage of the cell will decrease. Correct. There are two ways you can think about this: 1) LeChatelier's principle tells us that if you increase the concentration of product in an equilibrium reaction, you will cause the equilibrium to shift to the left. Shifting to the left in the case of a redox reaction means producing a lower potential. 2) The Q term in the Nernst equation is products over reactants. Increasing the ratio of products to reactants ([X2+] = 10 × [Y2+]) means the log(Q) term will become larger, but since that number is being subtracted from Eocell, Ecell is decreasing. In other words, increasing Q decreases Ecell, and decreasing Q increases Ecell.

A student constructs the following galvanic cell using a zinc electrode in 1.0 M Zn(NO3)2, a silver electrode in 1.0 M AgNO3, and a salt bridge containing aqueous KNO3. What is the cell notation for this electrochemical cell? A. Zn(s)│Zn2+(aq) || Ag+(aq)│Ag(s) B. Ag(s), Ag+(aq) || Zn2+(aq), Zn(s) C. Ag(s)│Ag+(aq) || KNO3(aq) || Zn2+(aq)│Zn(s) D. Ag(s)│Ag+(aq) || Zn2+(aq)│Zn(s)

A. Zn(s)│Zn2+(aq) || Ag+(aq)│Ag(s)

Consider the cell diagram below: Mg(s) | Mg2+(aq) || Au+(aq) | Au(s) Mg2+ + 2e-⟶ Mg, E° = -2.36 Au+ + e- ⟶ Au, E° = +1.69 What is the cathode and what is the cell type?

Au (s), a voltaic cell The left side of a cell diagram is always the anode (oxidation), and the right side of a cell diagram is always the cathode (reduction). Therefore, the cathode is Au+(aq) | Au(s). To determine the cell type E°cell = E°cathode - E° anode= +1.69 -(-2.36)= 4.05 v As the E°cell is positive, this is a voltaic cell (a battery).

2 Ga (l) + 3 Br2 (l) ⟶ 2 GaBr3 (s) In this reaction, what is the oxidizing agent and what is the reducing agent?

Br2 is the oxidizing agent Ga is the reducing agent 1. Assign oxidation numbers to each compound in the reaction. 2 Ga (l) + 3 Br2 (l) ⟶ 2 GaBr3 (s) 0 0 +3 -1 2. Compare oxidation numbers from reactant to products. Ga= 0 ⟶ Ga= +3, Gallium is oxidized, therefore it is the reducing agent since the electrons it loses reduced the charge of Bromine. Br= 0 ⟶ Br= -1, Bromine is reduced, therefore it is the oxidizing agent since the electrons it gains are a result of Gallium losing electrons.

A 3.7 amp current is passed through an electrolytic cell, and Al3+ is reduced to Al at the cathode. What mass of solid aluminum is produced after six hours? A. 1.24 g B. 0.828 g C. 7.45 g D. 22.3 g

C. 7.45 g Multiply current × time (I × t) to get the total charge in coulombs, and divide coulombs by 96,485 C/mol to get the number of moles of electrons flowing into the reaction. Then use the balanced half-reaction to calculate the number of moles of aluminum reduced from the number of moles of electrons. Moles of electrons = (3.7 amps × 6 hr × 60 min/hr × 60 sec/min)/96,485 C/mol = 0.8283 moles e- From the balanced half-reaction Al3+ + 3e- → Al, it takes 3 electrons to reduce one Al3+ ion to a metal atom, so moles of Al = 0.8283/3 = 0.276 mol. 0.276 mol × 26.98 g/mol = 7.45 g of Al.

Which species is the weakest reducing agent in the table (Links to an external site.) of half-reactions? A. Li+ B. Li C. F- D. F2

C. F- For a molecule to be a weak reducing agent, it has to be a bad oxidizer/good reducer. The reduction of F2 to F- has the largest positive value on the table of standard reduction values. Therefore, its reduction is the most spontaneous and it would be the worst reducing agent. For it to work as a reducing agent, F- would have to oxidize into F2. Therefore, F- is the weakest reducing agent.

In the decomposition of cobalt (II) chloride, what substance is being oxidized? CoCl2 (s) → Co (s) + Cl2 (g) CoCl2

Cl Chlorine went from an oxidation state of -1 to 0.

Balance this equation in acidic solution. What is the coefficient of cyanide ion? CN−(aq) + ClO2 (aq) ⟶ CNO−(aq) + Cl− (aq) A. 2 B. 1 C. 3 D. 5 E. 7

D. 5 1. Separate the overall equation into 2 half-reactions. CN−(aq) + ClO2 (aq) ⟶ CNO−(aq) + Cl− (aq) +2 -3 +4 -2 +2 -1 -2 -1 -1 +2 -1 -1 Oxidation half reaction = CN−(aq) ⟶ CNO−(aq) Reduction half reaction= ClO2 (aq) ⟶ Cl− (aq) 2. Balance each half reaction by balancing the charge using electrons. For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding a H+ ion to the side deficient in positive charge. 5 x (CN−(aq) ⟶ CNO−(aq) + 2e- + 2 H-)= 5 CN−(aq) ⟶ 5 CNO−(aq) + 10e- + 10 H- 2 x (ClO2 (aq)+ 5e- + 4 H- ⟶ Cl− (aq) ) = 2 ClO2 (aq)+ 10e- + 8 H- ⟶ 2Cl− (aq) 3. Balance the equation. 5 CN-(aq) + 2 ClO2 (aq) + H2O ⟶ 5 CNO (aq) + 2 Cl- (aq) + 2 H+

Using the standard potential tables, what is the largest approximate E° value that can be achieved when two half-cell reactions are combined to form a battery? A. 3 v B. -6 v C. -3 v D. 6 v

D. 6 v the largest values of E°red are about +3V and -3V. The species with the positive value would be reduced, the other would be oxidized. E°cell = E°cathode - E° anode= -3 -(-3)= 6 v

For the cell diagram below: Cd(s) | CdSO4(aq) || Hg2SO4 | Hg(l) | Pt(s) What reaction occurs at the cathode? A. 2 Hg(l) + SO4^2- (aq) ⟶ Hg2SO4 (s) + 2e- B. 2 Cd (l) + SO4^2- (aq) ⟶ CdSO4 (s) + 2 e- C. CdSO4(s) + 2e- ⟶ 2 Cd(l) + SO4^2- (aq) D. Hg2SO4 (s) + 2 e- ⟶ 2 Hg (l) + SO4^2- (aq)

D. Hg2SO4 (s) + 2 e- ⟶ 2 Hg (l) + SO4^2- (aq) The reaction at the cathode is ALWAYS a reduction. Therefore, electrons should appear on the reactant side of the half-reaction. The right side of the cell diagram shows the reduction half-reaction.

Which of the metals in the list below will react with 1M H2SO4 to produce hydrogen gas? Refer to the standard potentials list. Na+ + 1e- ⟶ Na E° = -2.714 Cd2+ + 2e- ⟶ Cd E° = -0.403 Pb2+ + 2e- ⟶ Pb E° = -0.126 Cu2+ + 2e- ⟶ Cu E° = +0.337 A. Na and Cd only B. Cu only C. Na, Cd, Pb, and Cu D. Na, Cd, and Pb only

D. Na, Cd, and Pb only For H2SO4 to produce hydrogen gas, it must be reduced. 2H+ + 2e- ⟶ H2 E° = 0.00 Therefore, the metals have to oxidize, or act as the anode. This process must be spontaneous (E°cell > 0). E°cell = E°cathode - E° anode= 0-E metal For Ecell to be positive, the E° of the metal reduction must be negative. As the reduction of copper has a positive E°, it is the only metal that will not spontaneously react with H2SO4 to make hydrogen gas.

When a piece of aluminum foil is dropped into an aqueous solution of FeCl2 in an electrochemical cell, the aluminum dissolves and solid iron is formed. 2Al(s) + 3Fe2+ (aq) → 2Al3+(aq) + 3Fe(s) You already know that in a spontaneous reaction the products have a lower free energy than the reactants do. In this reaction, where do the electrons have a lower free energy? A. When they are in one of the solutions B. When they are in the wire C. When they are in the solid aluminum D. When they are in solid iron

D. When they are in solid iron The electrons in an electrochemical cell will go to the lowest position available since this has the highest free energy. Since they leave the solid aluminum and end up in the solid iron, the electrons have a lower free energy in the iron atoms than they do in the aluminum atoms.

Silver is plated on copper by immersing a piece of copper into a solution containing silver (I) ions. In the plating reaction, copper... A. is reduced and is the oxidizing agent. B. is reduced and is the reducing agent. C. is oxidized and is the oxidizing agent. D. is oxidized and is the reducing agent.

D. is oxidized and is the reducing agent. This is a displacement reaction. The copper displaces the silver ions from solution, giving the reaction: Ag+ + Cu ⟶ Cu+ + Ag Copper loses electrons to form the ion, so copper is oxidized. Since copper gives electrons to silver, it cause silver to be reduced. Therefore, copper is oxidized and is the reducing agent.

Is this a galvanic cell or an electrolytic cell? What species is being produced at the cathode in this cell? Pt | H2 | H+ || Pb2+ | Pb Standard reduction potentials: 2H+ + 2e- → H2 E° = 0.00 V Pb2+ + 2e- → Pb E° = -0.13 V

Electrolytic, Pb You can tell from the shorthand notation that Pb is being produced because the anode is written on the left side and the cathode (reduction) is on the right side. That means the cathode reaction is Pb2+ + 2e- → Pb. Looking at the two half-cell potentials, the lead half-cell has a lower standard reduction potential than the hydrogen half-cell, so normally you would expect the lead half-cell to be the oxidation reaction. Since in this case it's written as the reduction half reaction, this is an electrolytic cell.

In the reaction between lead (II) sulfide and oxygen gas, what is the oxidizing agent? PbS (s) + 2O2 (g) → SO2 (g) + PbO2 (s)

O2

In Sn (s) + 2HCl (aq) ⟶ SnCl2 (aq) + H2 (g), what is being oxidized and what is being reduced?

Sn is being oxidized, H2 is being reduced. 1. Assign oxidation numbers to all atoms in the equation and ignore the coefficients in the equation. Sn (s) + 2HCl (aq) ⟶ SnCl2 (aq) + H2 (g) 0 +1 -1 +2 -1 0 2. Compare oxidation numbers from the reactant side to the product side of the equation. If a redox reaction has occurred, you will find that the oxidation numbers of two (no more/no less) elements have changed from the reactant side to the product side. The element oxidized is the one whose oxidation number increased. The element reduced is the one whose oxidation number decreased. Sn= 0 ⟶ Sn = +2, tin is oxidized H= +1 ⟶ H= 0=, hydrogen is reduced

True or false: In a working electrochemical cell (a galvanic cell or a battery), the cations in the salt bridge move toward the cathode.

True The cathode is ALWAYS the site of reduction. This means positive ions pick up electrons to become less positive. Therefore, the cathode is LOSING positive charge and cations (positively charged by definition) in the salt bridge will move towards the cathode to compensate.

In each half cell of an electrochemical cell, a half-reaction occurs. In the half-cell that the electrons are flowing out of, __________ occurs and in the half-cell that electrons are flowing into, ____________.

oxidation reduction