CH 20 & 21
What are the oxidation states (or charges) of potassium, chromium, and oxygen in K2CrO4? Enter the oxidation state for an individual atom of each element, separated by commas, in the order in which the elements appear in the compound (e.g., +1,+2,-3).
+1,+6,-2
Which of the following corresponds to the representation of a beta particle in a nuclear equation? 0−1e 00γ 42He 0+1e
0−1e A beta particle is a type of radiation that consists of high-speed electrons emitted by an unstable nucleus. See Section 21.1
For any that do not, describe a nuclear decay process that would alter the neutron-to-proton ratio in the direction of increased stability. 1 - β-decay 2 - positron emission or orbital electron capture 3 - α emission polonium-216 (Has more than 83 protons) chlorine-32 (Below belt of stability) neon-24 (Above belt of stability) tin-108 (Below belt of stability)
1 - β-decay: neon-24 2 - positron emission or orbital electron capture: chlorine-32 tin-108 3 - α emission: polonium-216
Give the symbol for the following. a proton
1 1H A proton can be represented as 1 1p or 1 1H.
In the process of oxidizing I− to I2, SO42− is reduced to SO2. How many moles of SO2 are produced in the formation of one mole of I2?
1 mol This question could be answered by looking at the coefficients in the full, balanced equation. However, it is easier to look at the change in the number of electrons for each half-reaction. Think of the electrons lost in the oxidation process as the fuel for the reduction process. For example, if A→B produces two electrons but X→Y consumes one electron, the X→Y reaction can occur twice for each instance of the A→B reaction. In other words, two moles of Y can be produced for each mole of A consumed. The iodine half-reaction generates two electrons. The sulfur half-reaction consumes two electrons. So we can determine that I2 and SO2 are produced in a 1:1 mole ratio without ever writing the full equation.
How does this unit relate to energy expressed in joules? (Regarding Volts) 1. A potential of one volt imparts ________ joule of energy to one _________ of charge.
1. A potential of one volt imparts *one* joule of energy to one *coulomb* of charge.
Nickel and aluminum electrodes are used to build a galvanic cell. The standard reduction potential for the nickel(II) ion is −0.26 V and that of the aluminum(III) ion is −1.66V. Complete the sentences describing the cell. (Answer with either Ni or Al) 1. In the nickel-aluminum galvanic cell, the cathode is ___________.
1. In the nickel-aluminum galvanic cell, the cathode is *Ni*.
Estimate the optimal number of neutrons for a nucleus of Yb (Z=70). https://session.masteringchemistry.com/problemAsset/3239002/3/MC_3239002_graph.png
102 neutrons (My answer of 103 was accepted)
Enter balanced nuclear equations for the following transformations. Nitrogen-13 undergoes electron capture.
13 7N + 0 −1e→13 6C Electron capture is the capture of an electron from the electron cloud of a surrounding proton-rich nuclide. This results in the decrease of the atomic number as a proton becomes a neutron and an electron is released.
When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation of two new nuclei and neutrons. The following equation is an example of one such fission process: 235 92U + 1 0n → AZBa + 94 36Kr + 3(1 0n) Enter the isotope symbol for the barium (Ba) nucleus in this reaction.
139 56Ba The uranium-235 isotope undergoes a neutron capture followed by a fission reaction. The first consideration is the type of nuclear decay. In this reaction the only decay is the splitting of the nucleus with the subsequent release of neutrons. Therefore, the number of protons does not change from the left side of the equation to the right. Use this as the starting point of your calculation, and keep in mind that the total number of nucleons on the right side of the equation must equal the total number of nucleons on the left.
The number of neutrons in the nuclide neptunium-237.
144 neutrons
Enter balanced nuclear equations for the following transformations. Gold-188 decays by positron emission.
188 79Au→ 0 +1e + 188 78Pt
Balance the following equation: K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O Generally coefficients of 1 are omitted from balanced chemical equations. When entering your answer, include coefficients of 1 as required for grading purposes. Enter the coefficients for each compound, separated by commas, in the order in which they appear in the equation (e.g., 1,2,3,4,5,6,7).
2,3,10,4,3,2,5 The balanced reaction will be 2K2CrO4 + 3Na2SO3 + 10HCl → 4KCl + 3Na2SO4 + 2CrCl3 + 5H2O First assign oxidation states to determine which atoms are being oxidized and which atoms are being reduced. Then, divide the equation into two, with one part representing the oxidation half-reaction and one part representing the reduction half-reaction. After balancing both half-reactions, multiply them by whole-number coefficients until all electrons cancel out when the sum of the half-reactions is found. In this case, once you have the balanced half-reactions you will need to add Cl−, Na+, and K+ ions and (sic) to both sides of the equation until all of the ions form neutral compounds. Finally, check to make sure that the total number of atoms of each element is the same on the right and left sides of the equation. Also check to make sure that the total charge on the left and right sides of the equation is the same. Identify the element that undergoes oxidation in the following reaction: K2CrO4+Na2SO3+HCl→KCl+Na2SO4+CrCl3+H2O To do so, take into account the oxidation state of each element. Consider that an increase in the oxidation state of an element from the reactant side to the product side indicates a loss of electrons by the element, or oxidation. (Answer is S) What is the oxidation half-reaction? SO32−+H2O→SO42−+2e−+2H+ Identify the element that undergoes reduction. To do so, take into account the oxidation state of each element. Consider that a decrease in the oxidation state of an element from the reactant side to the product side indicates a gain of electrons by the element, or reduction. (Answer is Cr) To balance the oxygen atoms of CrO42−, water is added to the reactant side of the equation and hydroxide ions are added to the product side of the equation. What is the reduction half-reaction for this equation? CrO42− + 8H+ + 3e− → Cr3+ + 4H2O To cancel the number of electrons on each side of the chemical equation, the total number of electrons should remain the same in both the oxidation and reduction half-reactions. What whole-number coefficients should the oxidation half-reaction and the reduction half-reaction be multiplied by to ensure that the number of electrons cancels out in the sum of the half-reactions? (3,2) What is the sum of these two half-reactions? Make sure that all electrons are cancelled out when the half-reactions are added together. 3SO32− + 2CrO42− + 10H+ → 3SO42− + 2Cr3+ + 5H2O Now, add Cl−, Na+, and K+ ions to both sides of the reaction to form neutral compounds on each side of the reaction. Recall that if you add five moles of Cl− to the product side of the reaction, you will also need to add five moles of Cl− to the reactant side.
Nickel and aluminum electrodes are used to build a galvanic cell. The standard reduction potential for the nickel(II) ion is −0.26 V and that of the aluminum(III) ion is −1.66V. Complete the sentences describing the cell. (Answer with either Ni or Al) 2. Therefore electrons flow from __________ to __________.
2. Therefore electrons flow from *Al* to *Ni*.
Enter balanced nuclear equations for the following transformations. Bismuth-213 undergoes alpha decay.
213 83Bi→ 4 2He + 209 81Tl Alpha decay of a nuclide yields an alpha particle (a helium atom) and a nuclide with a mass number that is 4 less than what it started with and an atomic number that is 2 less. As a result, the identity of the nuclide changes as well.
What substance is produced at the anode during the electrolysis of molten calcium bromide, CaBr2? Assume standard conditions. Express your answer as a chemical formula.
2Br−→Br2+2e− Molten calcium bromide contains calcium ions, Ca2+, and bromide ions, Br−. Therefore, the only possible reactions that can occur are Ca2+ + 2e− → Ca 2Br− → Br2 + 2e− Of the two half-reactions occurring in this cell, determine which one is an oxidation half-reaction and which one is a reduction half-reaction. The oxidation half-reaction occurs at the anode.
What substance is produced at the anode during the electrolysis of a mixture of molten calcium bromide, CaBr2(l), and molten magnesium iodide, MgI2(l)? Assume standard conditions. Express your answer as a chemical formula.
2I−→I2+2e− This molten mixture contains calcium ions, Ca2+, magnesium ions, Mg2+, bromide ions, Br−, and iodide ions, I−. Therefore, the possible reactions are Ca2+ + 2e− → Ca Mg2+ + 2e− → Mg 2Br− → Br2 + 2e− 2I− → I2 + 2e− You must now determine which of these reactions are oxidation half-reactions and which are reduction half-reactions. You must choose which of the two oxidation reactions will actually occur. Since oxidation is the loss of electrons, the less electronegative substance will take precedence at the anode. Because iodine lies below bromine in the periodic table, we know that it is less electronegative. Electronegativity is the tendency of an atom to draw electrons toward itself, so a lower electronegativity makes I− the better reducing agent (more likely to be oxidized). You must choose which of the two oxidation reactions will actually occur. The oxidation half-reaction with the greatest standard oxidation potential will take precedence at the anode. Consider the equations 2I− → I2 + 2e−; E∘ = −0.53V 2Br− → Br2 + 2e−; E∘ = −1.07V Since −0.53 is a greater (less negative) number than −1.07, we know that the iodine reaction takes precedence
Nickel and aluminum electrodes are used to build a galvanic cell. The standard reduction potential for the nickel(II) ion is −0.26 V and that of the aluminum(III) ion is −1.66V. Complete the sentences describing the cell. (Answer with either Ni or Al) 3. The __________ electrode loses mass, while the __________ electrode gains mass.
3. The *Al* electrode loses mass, while the *Ni* electrode gains mass. (Ni⁺², being the cathode, accepts electrons, becoming Ni(s) and depositing on the Ni electrodes.)
The number of protons in the nuclide 98 Mo
42 protons The number of protons is equal to the atomic number, which can be found by locating the element on the periodic table. The atomic number is usually given above the chemical symbol.
The number of protons in the nuclide 134 55Cs.
55 protons The atomic number, denoted by the subscript, is the number of protons in the nucleus.
The number of neutrons in the nuclide 98 Mo.
56 neutrons The superscript is the mass number, which is the number of protons and neutrons in the nucleus (in this case, 98). The number of neutrons is, therefore, the mass number minus the atomic number (98−42=56).
What is the missing product in the following nuclear reaction? Fe→_______+0−1e 5927Pr 5927Co 5927Fe 5925Mn
5927Co With the loss of the beta particle, the mass number does not change, but the atomic number increases by one. See Section 21.1
The number of neutrons in the nuclide 134 55Cs
79 neutrons The superscript is the mass number, which is the number of protons and neutrons in the nucleus (in this case, 134). The number of neutrons is, therefore, the mass number minus the number of protons (134−55=79).
Enter balanced nuclear equations for the following transformations. Rubidium-83 undergoes electron capture.
83 37Rb + 0 −1e→83 36Kr Electron capture is the capture of an electron from the electron cloud of a surrounding proton-rich nuclide. This results in the decrease of the atomic number as a proton becomes a neutron and an electron is released.
In another process in which 235 92U undergoes neutron bombardment, the following reaction occurs: 235 92U + 1 0n → AZSr + 143 54Xe + 3(1 0n) Enter the isotope symbol for the strontium (Sr) nucleus in this reaction.
90 38Sr The uranium-235 isotope undergoes a neutron capture followed by a fission reaction. The first consideration is the type of nuclear decay. In this reaction the only decay is the splitting of the nucleus with the subsequent release of neutrons. Therefore, the number of protons does not change from the left side of the equation to the right. Use this as the starting point of your calculation, and keep in mind that the total number of nucleons on the right side of the equation must equal the total number of nucleons on the left.
The number of protons in the nuclide neptunium-237.
93 protons
How many neutrons must an 56^26Fe nucleus capture to generate the unstable intermediate [73^26Fe] according to the equation: 5626Fe+ ? 10n→[7326Fe]
? = 17 Remember that in neutron capture, a neutron is taken up by the nucleus. The total number of neutrons captured is equal to the change in atomic mass between the starting nucleus and the product.
How many electrons must the unstable intermediate [73^26Fe] release to give the 73^32Ge nucleus according to the equation: [7326Fe]→7332Ge+ ? 0−1β
? = 6 Remember that in β decay, an electron is given off by a neutron resulting in the formation of a proton. Therefore, the total number of electrons given off equals the total number of protons formed.
230 90Th undergoes alpha decay. What is the mass number of the resulting element?
A = 226 The balanced nuclear equation is 230 90Th→226 88Ra+42He
234 91Pa undergoes beta decay. What is the mass number of the resulting element?
A = 234 The balanced nuclear equation is 234 91Pa→234 92U+ 0−1β
Uranium-238 decays by alpha emission. What is the mass number of the resulting element?
A = 234 The balanced nuclear equation is 238 92U→234 90Th+42He The total number of protons and neutrons must be the same on both sides of the equation. Decay by alpha emission means that an alpha particle will appear on the product side of the nuclear equation, and thus the mass number and atomic number should be subtracted from that of uranium-238 to determine the element resulting from the decay.
What is the value of A in the following nuclear reaction? 230 90Th→226 88Ra+AZX
A = 4 Balancing nuclear equations is slightly different than balancing chemical equations. The major difference is that in nuclear reactions we must account for protons, neutrons, and electrons, as well as write out the symbols for various chemical elements. In a nuclear equation, the products and reactants are symbolized as AZX where X is the chemical symbol for the element, A is the mass number, and Z is the atomic number. There are two main rules to remember when balancing nuclear equations: 1. The total of the superscripts (mass numbers, A) in the reactants and products must be the same. 2. The total of the subscripts (atomic numbers, Z) in the reactants and products must be the same.
For a spontaneous reaction A(aq)+B(aq)→A−(aq)+B+(aq) answer the following questions: If you made a voltaic cell out of this reaction, what half-reaction would be occurring at the cathode, and what half-reaction would be occurring at the anode? A(aq)+1e−→A−(aq) at the cathode and B(aq)→B+(aq)+1e− at the anode A(aq)+1e−→A−(aq) at the anode and B(aq)→B+(aq)+1e− at the cathode
A(aq)+1e−→A−(aq) at the cathode and B(aq)→B+(aq)+1e− at the anode
Electrolysis is performed on a mixture of CuBr(l), AgBr(l), MgBr2(l), and NiBr2(l). Which of the following is produced at the cathode?
Ag(s) When electricity is applied to a molten binary salt, the cation will be reduced and the anion will be oxidized. The electrolysis of CaBr2(l), for example, produces Ca(s) at the cathode (from the reduction of Ca2+) and Br2(l) at the anode (from the oxidation of Br−). If more than one cation is present, only the one with highest reduction potential will be reduced. Similarly, if more than one anion is present, only the one with the highest oxidation potential will be oxidized. The only anion present is Br−, making it the only choice for oxidation, which occurs at the anode. At the cathode, the cations will compete for reduction. Only the cation with the highest reduction potential will be reduced. Look at the reduction potentials for all the metal ions present and identify the one with the highest (most positive) value. The product of that reduction half-reaction will form at the cathode.
Consider a iron-silver voltaic cell that is constructed such that one half-cell consists of the iron, Fe, electrode immersed in a Fe(NO3)3 solution, and the other half-cell consists of the silver, Ag, electrode immersed in a AgNO3 solution. The two electrodes are connected by a copper wire. The Fe electrode acts as the anode, and the Ag electrode acts as the cathode. To maintain electric neutrality, you add a KNO3 salt bridge separating the two half-cells. Use this information to solve Parts B, C, and D. The half-cell is a chamber in the voltaic cell where one half-cell is the site of an oxidation reaction and the other half-cell is the site of a reduction reaction. Type the half-cell reaction that takes place at the cathode for the iron-silver voltaic cell. Indicate the physical states of atoms and ions using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not include phases for electrons. Express your answer as a chemical equation.
Ag+(aq)+e−→Ag(s)
Which electrode of a voltaic cell, the cathode or the anode, corresponds to the higher potential energy for the electrons?
Anode (Anode is the electrode made by the metal of lower reduction potential. So it has a lesser ability to get reduced by accepting electrons. So, it has a greater ability to lose electrons than the cathode. More electrons have been released at the anode as compared to the cathode. So, we may say that the anode is more able to "push" electrons to the other side because of this gradient of electrons created. That is why the anode has a higher electrode potential.)
In the context of the iron(II)-silver cell described in Part A, match each of the following descriptions to the anode or cathode. (Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)) Fe Ag gains mass loses mass attracts electrons positive electrode negative electrode stronger reducing agent
Anode: Fe negative electrode stronger reducing agent loses mass Cathode: Ag positive electrode attracts electrons gains mass The anode is where oxidation takes place. The cathode is where reduction takes place. Identify the reduction and oxidation reactions. (Fe/Fe2+) involves oxidation and (Ag+/Ag) involves reduction. The anode is where oxidation occurs and thus the anode loses electrons. The cathode is where reduction occurs and thus the cathode gains electrons. Electrons flow from anode to cathode. Consider the two half-reactions Fe2+(aq)+2e−→Fe(s) Ag+(aq)+e−→Ag(s) For which reaction is solid metal being produced? For which reaction is solid metal being consumed? How does this affect the mass of each solid electrode? The reaction in a galvanic cell is spontaneous, so the negative electrons naturally flow toward the positive electrode. Therefore, if you can identify which electrode is gaining electrons (reduction), you have identified the positive electrode. The stronger reducing agent is the metal that is more likely to be oxidized. Which electrode is oxidized? the anode
What type of nuclear decay involves the conversion of a neutron to a proton and the accompanying emission of high-speed electrons from the unstable nucleus?
Beta decay
Choose the one that is the stronger reducing agent. Ca(s) or Al(s). Ca = -2.87 Al = -1.66
Ca(s) (More negative number means it is the stronger *REDUCING* agent. Be aware that if you need the stronger oxidizing agent then you'll want the *less* negative number.)
What substance is produced at the cathode during the electrolysis of molten calcium bromide, CaBr2? Assume standard conditions. Express your answer as a chemical formula.
Ca2+ + 2e− → Ca Molten calcium bromide contains calcium ions, Ca2+, and bromide ions, Br−. Therefore, the only possible reactions that can occur are Ca2+ + 2e− → Ca 2Br− → Br2 + 2e− Of the two half-reactions occurring in this cell, determine which one is an oxidation half-reaction and which one is a reduction half-reaction. The reduction half-reaction occurs at the cathode.
For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C if the concentrations are [Co3+]= 0.756 M , [Co2+]= 0.348 M , and [Cl−]= 0.106 M , and the pressure of Cl2 is PCl2= 9.90 atm?
E = 0.416 V (E = E∘− ((2.303RT) / nF)logQ n = 2 Q = [products]^x / [reactants]^y) Because the potential is still a positive number, the reaction is spontaneous, even under these nonstandard conditions.
What is the cell potential for the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 61 ∘C when [Fe2+]= 3.60 M and [Mg2+]= 0.310 M. Express your answer to three significant figures and include the appropriate units. (E∘ = 1.92 V)
E = 1.96 V Use the Nernst equation E = E∘− ((2.303RT) / nF)logQ to calculate E. Plug in the values of E∘, T, n, and Q that you calculated in the previous parts.
You are given some ferrous iodide, FeI2, and are asked to extract the iron through electrolysis. What can be said about the merits of molten versus aqueous FeI2? You'll have to perform electrolysis on the molten salt because you can't get iron metal from electrolysis of the aqueous solution. You'll have to perform electrolysis of the aqueous solution because you can't get iron metal from electrolysis of the molten salt. Electrolysis of either the molten or aqueous salt will produce solid iron. Electrolysis will not produce solid iron regardless of whether the salt is molten or aqueous.
Electrolysis of either the molten or aqueous salt will produce solid iron. When current is applied to an aqueous solution of a binary salt, the reduction of water will compete with the reduction of the cation. Only the reduction reaction with the higher (more positive) potential will occur at the cathode. Here are the possible reduction reactions in CaBr2(aq) (Notice that this is not the compound from the question it is only an example of how to think about the problem): 2H2O(l)+2e− → H2(g)+2OH−(aq) E∘red=−0.83 V Ca2+(aq)+2e− → Ca(s) E∘red=−2.87 V Similarly, the oxidation of water will compete with the oxidation of the anion. Only the oxidation reaction with the higher (more positive) potential will occur at the anode. Here are the possible oxidation reactions in CuBr(aq): 2Br−(aq) → Br2(l)+2e− E∘ox=−1.07 V 2H2O(l) → O2+4H++4e− E∘ox=−1.23 V For the electrolysis of CaBr2(aq), the reduction of H2O and the oxidation of Br− are the favored half-reactions so you would see bubbles of hydrogen gas forming at the cathode instead of solid calcium like you would for the molten salt. If a metal ion is present in a solution and if the reduction of that ion is more favorable than that of water, the neutral metal will be produced during electrolysis of the solution. In other words, electrolysis of, say, a copper salt solution will produce copper but electrolysis of, say, a sodium salt solution will not produce sodium. Metals such as silver, copper, and nickel could also be produced through electrolysis of an aqueous solution of their salts because the reduction potentials of the metal ions are greater (more positive) than that of water. Metals such as magnesium, sodium, and calcium could only be produced through electrolysis of their molten salts because the reduction potentials of the metal ions are less (more negative) than that of water.
Calculate the standard cell potential (E∘) for the reaction X(s)+Y+(aq)→X+(aq)+Y(s) if K = 7.12×10−3. Express your answer to three significant figures and include the appropriate units.
E∘ = -0.127 V First, determine the value of n. Then, use the following equation to find E∘: lnK = nFE∘ / RT Keep in mind that standard temperature is 298 K. When E∘<0 and K<1 the reaction favors reactants.
Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 61 ∘C , where [Fe2+] = 3.60 M and [Mg2+] = 0.310 M. Calculate the standard cell potential at 25 ∘C for Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) Express your answer to three significant figures and include the appropriate units. (Fe2+(aq)+2e−→Fe(s); −0.45 Mg2+(aq)+2e−→Mg(s); −2.37)
E∘ = 1.92 V (Fe2+(aq)+2e−→Fe(s); −0.45 Mg2+(aq)+2e−→Mg(s); −2.37 E∘cell = E∘cathode − E∘anode)
The standard reduction potentials of lithium metal and chlorine gas are as follows: Li+(aq)+e−→Li(s); E∘cell = −3.04V Cl2(g)+2e−→2Cl−(aq); E∘cell = +1.36 In a galvanic cell, the two half-reactions combine to 2Li(s)+Cl2(g)→2Li+(aq)+2Cl−(aq) Calculate the cell potential of this reaction under standard reaction conditions.
E∘ = 4.40 V (E∘cell = E∘cathode − E∘anode)
Tarnish on iron is the compound FeO. A tarnished iron plate is placed in an aluminum pan of boiling water. When enough salt is added so that the solution conducts electricity, the tarnish disappears. Imagine that the two halves of this redox reaction were separated and connected with a wire and a salt bridge. Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Fe2++2e−→Fe;E∘=−0.440 V Express your answer to two decimal places and include the appropriate units.
E∘cell = 1.22 V The aluminum pan is the anode and the iron plate is the cathode.
Calculate the standard potential for the following galvanic cell: Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s) which has the overall balanced equation: Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s) Express your answer to three significant figures and include the appropriate units. Ag+(aq)+e−→Ag(s) E∘ = 0.80V Cu2+(aq)+2e−→Cu(s) E∘ = 0.34V Ni2+(aq)+2e−→Ni(s) E∘ = −0.26V Fe2+(aq)+2e−→Fe(s) E∘ = −0.45V Zn2+(aq)+2e−→Zn(s) E∘ = −0.76V
E∘cell = 1.25 V First, identify which half-reactions take place. Next, use the table in the introduction to determine the standard reduction potential for each of those half-reactions. Then calculate the standard cell potential as defined below. E∘cell=E∘red(cathode)−E∘red(anode)
Nickel and aluminum electrodes are used to build a galvanic cell. The standard reduction potential for the nickel(II) ion is −0.26 V and that of the aluminum(III) ion is −1.66V. What is the theoretical cell potential assuming standard conditions?
E∘cell = 1.40 V (Determine which half reaction will be the oxidation and the reduction reactions. Oxidation occurs at anode and reduction occurs at cathode. E∘cell = E∘cathode − E∘anode Also recall that spontaneous redox reactions must have a positive sign for E∘cell.)
For a spontaneous reaction A(aq)+B(aq)→A−(aq)+B+(aq) answer the following questions: What is the sign of E∘cell?
E∘cell is positive. (It says that the reaction is spontaneous which means that E∘cell must be positive.)
Consider a iron-silver voltaic cell that is constructed such that one half-cell consists of the iron, Fe, electrode immersed in a Fe(NO3)3 solution, and the other half-cell consists of the silver, Ag, electrode immersed in a AgNO3 solution. The two electrodes are connected by a copper wire. The Fe electrode acts as the anode, and the Ag electrode acts as the cathode. To maintain electric neutrality, you add a KNO3 salt bridge separating the two half-cells. Use this information to solve Parts B, C, and D. What is the net cell reaction for the iron-silver voltaic cell? Express your answer as a chemical equation.
Fe(s)+3Ag+(aq)→Fe3+(aq)+3Ag(s) The net cell reaction is derived by adding the anode half-cell reaction and the cathode half-cell reaction. In the iron-silver voltaic cell, at the anode three electrons are lost by the one Fe(s) atom, and, at the cathode, one electron is gained by one silver ion. The oxidation half-cell reaction and reduction half-cell reaction for a iron-silver voltaic cell are the following: anode (oxidation half-cell reaction): Fe(s)→Fe3+(aq)+3e− (1) cathode (reduction half-cell reaction): Ag+(aq)+e−→Ag(s) (2) Thus, the cathode half-cell reaction should be multiplied by a factor of 3 to equate the number of electrons transferred. Therefore, the cathode half-cell reaction will be 3Ag+(aq)+3e−→3Ag(s) (3) Now you can add reactions (1) and (3) to get the net cell reaction for the iron-silver voltaic cell.
Consider a iron-silver voltaic cell that is constructed such that one half-cell consists of the iron, Fe, electrode immersed in a Fe(NO3)3 solution, and the other half-cell consists of the silver, Ag, electrode immersed in a AgNO3 solution. The two electrodes are connected by a copper wire. The Fe electrode acts as the anode, and the Ag electrode acts as the cathode. To maintain electric neutrality, you add a KNO3 salt bridge separating the two half-cells. Use this information to solve Parts B, C, and D. The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell is the site of the reduction reaction. Type the half-cell reaction that takes place at the anode for the iron-silver voltaic cell. Indicate the physical states of atoms and ions using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not include phases for electrons. Express your answer as a chemical equation.
Fe(s)→Fe3+(aq)+3e−
Au3+(aq)+3e−→Au(s); +1.50 V Ag+(aq)+e− →Ag(s); +0.80 V Cu2+(aq)+2e−→Cu(s); +0.34 V Fe3+(aq)+3e−→Fe(s); −0.04 V Pb2+(aq)+2e−→Pb(s); −0.13 V Sn2+(aq)+2e−→Sn(s); −0.14 V Fe2+(aq)+2e−→Fe(s); −0.44 V Zn2+(aq)+2e−→Zn(s); −0.76 V Al3+(aq)+3e−→Al(s); −1.66 V Mg2+(aq)+2e−→Mg(s); −2.38 V Based on their standard oxidation potentials, rank these metals from most reactive to least reactive. gold aluminum tin iron
From most to least reactive reactive aluminum iron tin gold (The list consists of the reduction potentials and it wants them ordered based on oxidation potentials. To turn the reduction list into an oxidation list, simply swap the positive/negative values.)
Choose the one that is the stronger reducing agent. H2 (g, acidic solution) or H2S(g). H2 = 0.00? H2S = +0.14
H2 (g, acidic solution) (More negative number means it is the stronger *REDUCING* agent. Be aware that if you need the stronger oxidizing agent then you'll want the *less* negative number.)
Indicate whether the following balanced equations involve oxidation-reduction. Check all that apply. H2SO4(aq)+6NO2(g)+2H2O(l)→S(s)+6HNO3(aq) PBr3(l)+3H2O(l)→H3PO3(aq)+3HBr(aq) NaI(aq)+3HOCl(aq)→NaIO3(aq)+3HCl(aq)
H2SO4(aq)+6NO2(g)+2H2O(l)→S(s)+6HNO3(aq) NaI(aq)+3HOCl(aq)→NaIO3(aq)+3HCl(aq) Assign oxidation numbers to all of the atoms to determine if oxidation and reduction is taking place. In the reaction involving PBr3, none of the oxidation numbers change, so it is not considered a redox reaction. For the reactions involving NaI and NO2, some oxidation numbers change over the course of the reaction, so oxidation-reduction is taking place.
Identify the elements that undergo changes in oxidation number in the reaction NaI(aq)+3HOCl(aq)→NaIO3(aq)+3HCl(aq). Check all that apply. H I O Na Cl
I Cl In this reaction, I is oxidized from −1 to +5, while Cl is reduced from +1 to −1.
Choose the one that is the stronger reducing agent. BrO−3(aq) or IO−3(aq). BrO−3(aq) = +1.52 IO−3(aq) = +1.20
IO−3(aq) (More negative number means it is the stronger *REDUCING* agent. Be aware that if you need the stronger oxidizing agent then you'll want the *less* negative number.)
Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ∘C) for the following reaction: Fe(s)+Ni2+(aq)→Fe2+(aq)+Ni(s) Ni2+(aq)+2e−→Ni(s); −0.26 Fe2+(aq)+2e−→Fe(s); −0.45
K = 2.68×10^6 The equilibrium constant, K, for a redox reaction is related to the standard potential, E∘, by the equation lnK = nFE∘ / RT where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e−) , R (the gas constant) is equal to 8.314 J/(mol⋅K) , and T is the Kelvin temperature. When E∘>0 and K>1 the reaction favors the products.
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are Cu2+(aq)+2e−→Cu(s) and Fe(s)→Fe2+(aq)+2e− The net reaction is Cu2+(aq)+Fe(s)→Cu(s)+Fe2+(aq) Use the given standard reduction potentials in your calculation as appropriate. https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_keq_gibb/index.html E∘Cu 0.337 V E∘Fe -0.440 V R 8.314 J⋅mol−1⋅K−1 F 96,485 C/mol T 298 K
Keq = 1.92×10^26 First, identify the metal that acts as the cathode and the metal that acts as the anode in the given reaction using the two half-reactions that occur in the cell. Once you have identified the metal that acts as the cathode and the metal that acts as the anode, calculate the standard reduction potential E∘cell by using the following relation: E∘cell=E∘cathode−E∘anode Then, determine n (the number of electrons transferred in the given redox reaction) using the two half-reactions that occur in the cell. Finally, using the interactive activity, identify the relation between E∘cell and Keq. Rearrange the equation to calculate the equilibrium constant (Keq).
Label the diagram according to the components and processes of a voltaic cell.
Left side from top to bottom: Flow of electrons - This target refers to an arrow directed to the right below the measuring device. Flow of anions - This target refers to an arrow directed to the left inside the salt bridge. Zinc Anode - This target refers to the left electrode. Oxidation half-cell - This target refers to the left beaker. Right side from top to bottom: Flow of cations - This target refers to an arrow directed to the right inside the salt bridge. Copper Cathode - This target refers to the right electrode. Reduction half-cell - This target refers to the right beaker. In a voltaic cell, the electrons flow through an external circuit from the anode to the cathode. As the electrons moves toward the cathode, negative charge accumulates at the cathode. To neutralize the accumulation of negative charge at the cathode, the positive ions (cations) within the salt bridge flow toward the cathode. When electrons move toward the cathode, the positive charge accumulates at the anode and, to neutralize the accumulated positive charge, the negative ions (anions) within the salt bridge flow toward the anode.
Choose the one that is the stronger reducing agent. Fe(s) or Mg(s). Fe = -0.44 Mg = -2.37
Mg(s) (More negative number means it is the stronger *REDUCING* agent. Be aware that if you need the stronger oxidizing agent then you'll want the *less* negative number.)
What substance is produced at the cathode during the electrolysis of a mixture of molten calcium bromide, CaBr2(l), and molten magnesium iodide, MgI2(l)? Assume standard conditions. Express your answer as a chemical formula.
Mg2++2e−→Mg This molten mixture contains calcium ions, Ca2+, magnesium ions, Mg2+, bromide ions, Br−, and iodide ions, I−. Therefore, the possible reactions are Ca2+ + 2e− → Ca Mg2+ + 2e− → Mg 2Br− → Br2 + 2e− 2I− → I2 + 2e− You must now determine which of these reactions are oxidation half-reactions and which are reduction half-reactions. You must choose which of the two reduction reactions will actually occur. Since reduction is the gain of electrons, the more electronegative substance will take precedence at the cathode. Because magnesium lies above calcium in the periodic table, we know that it is more electronegative. Electronegativity is the tendency of an atom to draw electrons toward itself, so a higher electronegativity makes Mg2+ the better oxidizing agent (more likely to be reduced). You must choose which of the two reduction reactions will actually occur. The reduction half-reaction with the greatest (least negative) standard reduction potential will take precedence at the cathode. Consider the equations Mg2++2e−→Mg;E∘=−2.37V Ca2++2e−→Ca;E∘=−2.87V Since −2.37 is a greater (less negative) number than −2.87, therefore we know that the magnesium reaction takes precedence.
Identify the elements that undergo changes in oxidation number in the reaction H2SO4(aq)+6NO2(g)+2H2O(l)→S(s)+6HNO3(aq). Check all that apply. H O N S
N S In this reaction, S is reduced from +6 to 0, while N is oxidized from +4 to +5.
For which of the following aqueous salts will electrolysis produce hydrogen gas and oxygen gas? NaF(aq) NiI2(aq) NiF2(aq) CuBr(aq) NaI(aq)
NaF(aq) Hydrogen is produced from the reduction of water. Oxygen is produced from the oxidation of water. You need to identify the salts for which water is favored for both the reduction and oxidation half reactions. In other words, the cation of the salt must have a lower reduction potential than water and the anion of the salt must have a lower oxidation potential than water.
In the absence of an electric field, a radioactive beam strikes a fluorescent screen at a single point. When an electric field is applied, the radioactive beam is separated into three different components. One of the components is deflected toward the positive electrode because it is negatively charged, one of the components is deflected toward the negative electrode because it is positively charged, and one component is not deflected in any direction; instead, it moves along a straight path. *Identify the charges possessed by the different components of the radioactive beam by observing their behavior under the influence of an electric field.* γ rays α rays β rays
Negatively charged: β rays Neutral: γ rays Positively charged: α rays Under the influence of an electric field, α rays get deflected toward the negative electrode, indicating that they carry a positive charge. β rays get deflected toward the positive electrode, indicating that they are negatively charged. γ rays take a straight pathway, which shows that they are neutral. α rays are deflected to a lesser extent than β rays, indicating that α rays have a higher mass.
For the given reactions, classify the reactants as the reducing agent, oxidizing agent, or neither. F2 + H2 → 2HF 2Mg + O2 → 2MgO
Oxidizing agent: F2 O2 Reducing agent: Mg H2
Here is a more complex redox reaction involving the permanganate ion in acidic solution: 5Fe2+ + 8H+ + MnO4− → 5Fe3+ +Mn2+ + 4H2O Classify each reactant as the reducing agent, oxidizing agent, or neither. H+ Fe2+ MnO4-
Oxidizing agent: MnO4− Reducing agent: Fe2+ Neither: H+
https://session.masteringchemistry.com/problemAsset/3239035/3/Brown12e.ch21.p22.jpg Which of the following statements best explains the sawtooth variation across the series? The elements with an odd atomic number lie below the belt of stability. The elements with an odd atomic number lie above the belt of stability. Pairs of protons have a special stability, so elements with even atomic numbers have more abundant isotopes. The elements with an even atomic number have a magic number of protons.
Pairs of protons have a special stability, so elements with even atomic numbers have more abundant isotopes. Evidence shows that pairs of protons and neutrons have special stability, so elements with even atomic numbers have more abundant isotopes.
Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 61 ∘C , where [Fe2+] = 3.60 M and [Mg2+] = 0.310 M. What is the value for the reaction quotient, Q, for the cell?
Q = 8.61×10^−2 (Notice that the concentrations are different by a magnitude) In the reaction quotient, the concentrations of the aqueous products appear in the numerator, and the concentrations of the aqueous reactants appear in the denominator. Just like an equilibrium constant, Kc, the coefficients in the balanced equation must also be considered. Q = [Mg2+] / [Fe2+]
What final product is formed when thorium-230 undergoes a series of two alpha (α) emissions?
Radon-222
If you complete and balance the following equation in acidic solution Mn2+(aq) + NaBiO3(s) → Bi3+(aq) + MnO4−(aq) + Na+(aq) how many water molecules are there in the balanced equation (for the reaction balanced with the smallest whole-number coefficients)?
Seven on the product side The balanced equation for the reaction in acidic solution is 14 H+(aq) + 2 Mn2+(aq) + 5 NaBiO3(s) → 2 MnO4−(aq) + 5 Bi3+(aq) + 5 Na+(aq) + 7 H2O(l) There are 7 water molecules on the product side. https://mediaplayer.pearsoncmg.com/assets/_video.true/secs-blbmws-interactive-sample-exercise-20-2
Classify each nucleus as stable or unstable. 8942Mo 212 84Po 118 50Sn 3015P
Stable: 118 50Sn Unstable: 212 84Po 8942Mo 3015P Predicting nuclear stability is important when determining whether or not a nuclear reaction can take place spontaneously. Use the following set of guidelines to predict the nuclear stability of the elements listed below. 1. Stable nuclei have a high neutron-to-proton ratio, greater than 1.25 beyond atomic number 40 and continuing to rise at higher atomic numbers. 2. The majority of stable nuclei have even numbers of neutrons and protons. Odd-even and even-odd combinations can be stable but stability is less likely. 3. Nuclei that have the magic numbers of 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50, 82 or 126 neutrons are more stable than nuclei that do not contain these numbers. 4. Nuclei that have atomic numbers greater than 83 are unstable. Nuclei are considered stable if one or more criteria are met. (Sn - The N:P ratio is 1.36 which is higher than 1.25. and its atomic number is semi close to 40 so this is stable. Po - has an atomic number greater than 83 so it is unstable. Mo - The N:P ratio is only 1.12 and it should be above 1.25 to be stable. P - Using the rules given it's not really possible to tell if this should be stable or not. In actuality the only stable form of P is 31P and this is 30P which means that 30P is unstable.)
α and β rays consist of particles, whereas γ rays are high-energy radiation and contain no particles and thus they have no mass. Given that the penetrating power decreases with mass, rank the rays according to their penetrating power. Rank the components from strongest penetrating power to weakest penetrating power. To rank items as equivalent, overlap them. γ rays α rays β rays
Strongest to Weakest γ rays β rays α rays
Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 61 ∘C , where [Fe2+] = 3.60 M and [Mg2+] = 0.310 M. What is the value for the temperature, T, in kelvins? Express your answer to three significant figures and include the appropriate units.
T = 334 K
For a spontaneous reaction A(aq)+B(aq)→A−(aq)+B+(aq) answer the following questions: Which half-reaction from part above is higher in potential energy? The anode reaction is higher in potential energy. The cathode reaction is higher in potential energy.
The anode reaction is higher in potential energy. (The potential energy of electrons is higher in the anode than in the cathode. Thus, electrons flow spontaneously toward the electrode with the more positive electrical potential.)
An old iron beam was coated with an unknown metal. There is a crack on the coating and it is observed that the iron is rusting at the fracture. The beam is in a structure that experiences high stress, resulting in frequent fractures to the coating. What was the old metal coating likely made of and what metal you would use to repair the fractures to avoid further corrosion? Because rusting is observed at the fracture, what can be said about the reactivities of the iron and its coating? Neither the coating nor the iron is reactive. The iron is more reactive than the coating. The coating is more reactive than the iron.
The iron is more reactive than the coating. A more reactive metal would have prevented rusting, even at a fracture. Thus, you can rule out any metals that are more reactive than iron as possibilities for the original coating. If there is more than one choice left, consider cost as a factor.
The standard reduction potentials of lithium metal and chlorine gas are as follows: Li+(aq)+e−→Li(s); E∘cell = −3.04V Cl2(g)+2e−→2Cl−(aq); E∘cell = +1.36 In a galvanic cell, the two half-reactions combine to 2Li(s)+Cl2(g)→2Li+(aq)+2Cl−(aq) What can be said about the spontaneity of this reaction?
The reaction is spontaneous as written. (E∘ = 4.40 V and therefore ΔG∘ = -849 kJ. As ΔG∘ is negative it is spontaneous as written.)
At 900∘C titanium tetrachloride vapor reacts with molten magnesium metal to form solid titanium metal and molten magnesium chloride. Write a balanced equation for this reaction. Express your answer as a chemical equation. Identify all of the phases in your answer.
TiCl4(g)+2Mg(l)→Ti(s)+2MgCl2(l) First, the formulas for the reactants and products are determined. The name of titanium tetrachloride implies that there are four chlorine atoms for every titanium atom. Magnesium chloride has the formula MgCl2 because magnesium forms 2+ cations and chlorine forms 1− anions, and the overall molecule is neutral. Once the equation is written, a coefficient of 2 is placed in front of magnesium chloride to balance the chlorine atoms and, subsequently, in front of the molten magnesium to balance the magnesium.
What are the units for electrical potential?
Volt
Cathodic protection of iron involves using another more reactive metal as a sacrificial anode. Classify each of the following metals by whether they would or would not act as a sacrificial anode to iron under standard conditions. (compare to Fe2+ not Fe3+) Pb Sn Zn Au Ag Cu Mg Au3+(aq)+3e−→Au(s); +1.50 V Ag+(aq)+e− →Ag(s); +0.80 V Cu2+(aq)+2e−→Cu(s); +0.34 V Fe3+(aq)+3e−→Fe(s); −0.04 V Pb2+(aq)+2e−→Pb(s); −0.13 V Sn2+(aq)+2e−→Sn(s); −0.14 V Fe2+(aq)+2e−→Fe(s); −0.44 V Zn2+(aq)+2e−→Zn(s); −0.76 V Al3+(aq)+3e−→Al(s); −1.66 V Mg2+(aq)+2e−→Mg(s); −2.38 V
Will act as sacrificial anode for iron: Zn Mg Will not act as sacrificial anode for iron: Pb Sn Au Ag Cu For cathodic protection to occur, the more reactive metal must be in contact with the iron but does not need to coat it. However, the sacrificial anode will eventually corrode away or develop its own protective layer, leaving the iron vulnerable. By coating iron in zinc (a process called galvanizing) the protective layer that develops on the zinc will prevent further corrosion of both the zinc and the iron. Magnesium is cheaper than zinc, but it does not develop its own protective layer like zinc does. For this reason, magnesium is more commonly used in true sacrificial anode situations whereas zinc is used for galvanizing.
What chemical symbol should be used for X in the following nuclear reaction? 230 90Th→226 88Ra+AZX
X = He The complete designation for the product, an alpha particle, is 42He. An alternate designation frequently used is 42α or simply α.
What is the value of Z in the following nuclear reaction? 230 90Th→226 88Ra+AZX
Z = 2 Balancing nuclear equations is slightly different than balancing chemical equations. The major difference is that in nuclear reactions we must account for protons, neutrons, and electrons, as well as write out the symbols for various chemical elements. In a nuclear equation, the products and reactants are symbolized as AZX where X is the chemical symbol for the element, A is the mass number, and Z is the atomic number. There are two main rules to remember when balancing nuclear equations: 1. The total of the superscripts (mass numbers, A) in the reactants and products must be the same. 2. The total of the subscripts (atomic numbers, Z) in the reactants and products must be the same.
234 92U undergoes alpha decay. What is the atomic number of the resulting element?
Z = 90 The balanced nuclear equation is 234 92U→230 90Th+42He
234 90Th undergoes beta decay. What is the atomic number of the resulting element?
Z = 91 The balanced nuclear equation is 234 90Th→234 91Pa+ 0−1β Consider that during beta decay, a proton turns into a neutron and an electron. Thus, the beta particle is seen as having "negative one" protons.
In order to protect underground pipelines and storage tanks made of iron, the iron is made the cathode of a voltaic cell through a process called cathodic protection galvanization rusting sacrificial anodization
cathodic protection (Cathodic protection is the protecting of a metal from corrosion by making it the cathode in an electrochemical cell.)
Indicate whether each of the following nuclides lies within the belt of stability in Figure 21.3 in the textbook. chlorine-32 neon-24 tin-108 polonium-216
in the belt of stability: (None) out of the belt of stability: chlorine-32 neon-24 tin-108 polonium-216
Metal plating is done by passing current through a metal solution. For example, an item can become gold plated by attaching the item to a power source and submerging it into a Au3+ solution. The item itself serves as the cathode, at which the Au3+ ions are reduced to Au(s). A piece of solid gold is used as the anode and is also connected to the power source, thus completing the circuit. What mass of gold is produced when 8.50 A of current are passed through a gold solution for 23.0 min?
mass of Au = 7.98 g Using 1A = 1C/s, determine how many coulombs pass through this solution in 23.0 min. 1.17×10^4 C Based on the amount of current passed, how many moles of electrons pass through this solution? Recall that 1 mol of electrons=96,500C. 0.122 mol Based on the number of moles of electrons that pass through this solution, how many moles of gold are reduced? 4.05×10^−2 mol (Convert the moles of gold to grams.)
Give the symbol for a neutron.
n The symbol for a neutron is 10n or n. The subscript 0 indicates the neutron charge, and the superscript 1 indicates its mass. A neutron is an electrically neutral particle and has a mass of 1.0087 amu.
Consider the reaction Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s) at 61 ∘C , where [Fe2+] = 3.60 M and [Mg2+] = 0.310 M. What is the value for n?
n = 2 mol The value for n is equal to the number of moles of electrons transferred in the balanced overall reaction.
What is being oxidized, and what is being reduced? Express your answers as chemical formulas separated by a comma. (TiCl4(g)+2Mg(l)→Ti(s)+2MgCl2(l))
oxidized, reduced Mg,TiCl4 Recall that oxidation is the loss of electrons and reduction is the gain of electrons. In this reaction, magnesium goes from an oxidation number of 0 to +2, implying a loss of two electrons, while titanium goes from +4 to 0, implying a gain of four electrons.
Which substance is the reductant and which is the oxidant? Express your answers as chemical formulas separated by a comma. (TiCl4(g)+2Mg(l)→Ti(s)+2MgCl2(l))
reductant, oxidant Mg,TiCl4 The oxidant is a substance that oxidizes another substance. It acquires electrons and so is reduced. The substance that gives up electrons is the reductant, which reduces another substance and is oxidized itself.
In Part B you calculated ΔG∘ for the following redox reaction: Cu2+(aq)+Fe(s)→Cu(s)+Fe2+(aq), ΔG∘ = −1.50×10^5J Based on the value of ΔG∘ for the given redox reaction, identify the spontaneity of the reaction.
spontaneous When ΔG∘ is negative, the reaction is spontaneous. The ΔG∘ value for the redox reaction, Cu2+(aq)+Fe(s)→Cu(s)+Fe2+(aq), is −1.50×10^5J; hence, the reaction is spontaneous.
An old iron beam was coated with an unknown metal. There is a crack on the coating and it is observed that the iron is rusting at the fracture. The beam is in a structure that experiences high stress, resulting in frequent fractures to the coating. What was the old metal coating likely made of and what metal you would use to repair the fractures to avoid further corrosion? The old coating was likely made of __________. _________ would be a good choice for repairing the fracture. tin gold aluminum
tin aluminum In a spontaneous redox reaction involving two metal electrodes, the anode will always corrode while the cathode is always inert. The substance being reduced in these processes is not iron (the cathode) but other ions that come in contact with the iron.
The standard reduction potentials of lithium metal and chlorine gas are as follows: Li+(aq)+e−→Li(s); E∘cell = −3.04V Cl2(g)+2e−→2Cl−(aq); E∘cell = +1.36 In a galvanic cell, the two half-reactions combine to 2Li(s)+Cl2(g)→2Li+(aq)+2Cl−(aq) Calculate the free energy ΔG∘ of the reaction.
ΔG∘ = -849 kJ Use the formula ΔG∘=−nFE∘cell where n is the number of moles of electrons transferred and F=96,500J/V⋅mol e− is the Faraday constant.
In the activity, click on the Keq and ΔG∘ quantities to observe how they are related. Calculate ΔG∘ using this relationship and the equilibrium constant (Keq) obtained in Part A at T=298K: Keq = 1.92×10^26 Express the Gibbs free energy (ΔG∘) in joules to three significant figures. https://media.pearsoncmg.com/bc/bc_0media_chem/legacy_media/int_act_keq_gibb/index.html
ΔG∘ = −1.50×10^5 J ΔG∘ = −RT lnKeq The Gibbs free energy (ΔG∘) can also be calculated using the equation ΔG∘=−nFE∘cell. Recall that the number of electrons transferred (n) is 2, Faraday's constant (F) is 96,485 C/mol, and the E∘cell value obtained in Part A is 0.777 V. Substituting these values into the equation will give ΔG∘ = −2×96,485×0.777 = −1.50×10^5 Hence, ΔG∘ can be calculated by using both equations, and the answer will be −1.50×10^5 in either case.
Give the symbol for an alpha particle.
α Alpha particles are helium-4 nuclei. The symbol for an alpha particle is 42He or simply α. The left subscript, 2, is the atomic number. The atomic number can be found in the periodic table. The left superscript, 4, is the mass number. The mass number is equal to the sum of the protons and neutrons in the atom.
Given below are statements that summarize the characteristics of α, β, and γ rays. Identify the characteristics that correspond to each type of radiation. It is a high-speed electron. It is symbolized as 0−1e. It possesses neither mass nor charge. It has the weakest penetrating power. It is symbolized as 42He. It has the strongest penetrating power. It is the most massive of all the components.
α rays: It has the weakest penetrating power. It is symbolized as 42He. It is the most massive of all the components. β rays: It is a high-speed electron. It is symbolized as 0−1e. γ rays: It possesses neither mass nor charge. It has the strongest penetrating power. The three main components of a radioactive beam are α, β, and γ rays. Of these components, α rays contain positively charged particles known as α particles, β rays contain negatively charged particles known as β particles, and γ rays are electromagnetic radiation. Experimental evidence has shown that α rays are helium nuclei, which are represented as 42He, that β rays are high-energy electrons, represented as 0−1e, and that γ rays are photons with very high energy and no mass. α rays are the most massive among the three. It is evident from the sizes of the components that α rays have the lowest penetrating power and that γ rays possess the highest penetrating power. These characteristics are useful in studying radioactivity phenomena. In the process of natural radioactivity, α, β, or γ radiation are spontaneously emitted. The process can be called α decay, β decay, or γ decay, depending upon the type of emission involved.
Give the symbol for the following. a positron
β+ A positron can be represented as 0 +1e, e+, or β+.
Give the symbol for the following. a beta particle
β− A beta particle can be represented as 0 −1e, e−, or β−.
Give the symbol for gamma radiation.
γ Gamma radiation (or gamma rays) consists of high-energy photons. The symbol for gamma radiation is γ. The particle has no mass or charge, and if any superscript or subscript is given before the symbol, they are written as 0.
