Answer the four questions below regarding a particular type of alkaline battery in which the following half-reactions are used:

Question

Answer the four questions below regarding a particular type of alkaline battery in which the following half-reactions are used:

2MnO₂ ​(s) + H₂​O(l) + 2e^- Mn₂​O₃ ​(s) + 2OH^- (aq) E° = 0.150 V

Zn²⁺(aq) + 2e^- Zn(s) E° = −0.760 V

A. Balanced Equation for spontaneous redox reaction

B. Calculate the Standard Cell Potential

C. If the initial concentrations of Zn²⁺ and of OH^– are 1.0 × 10^–7 and 2.0 × 10^–7, respectively, calculate the cell potential after the Zn²⁺ concentration has increased by a factor of 10.0. 

D. At this point, what percentage of the initial voltage remains?

Answer 1

1. Zinc having the more negative potential between these two reactions implies that the zinc half reaction is more likely to be oxidation, and the manganese reaction is more likely to be a reduction (positive potential). This is the case if a spontaneous (galvanic) reaction was to occur. Therefore, the half-reactions can be written as such:
2. Zn(s) Zn²⁺(aq) + 2e E° = −0.760 V [electron donor, oxidation]
3. 2MnO₂ ​(s) + H₂​O(l) + 2e^- Mn₂​O₃ ​(s) + 2OH^- (aq) E° = 0.150 V [electron acceptor, reduction]
4. Overall reaction (electrons cancel out): 2MnO₂ ​(s) + H₂​O(l) + Zn(s) Zn²⁺(aq) + Mn₂​O₃ ​(s) + 2OH^-
5. Note that half-cell potentials are reversible, so the same potential in volts can be used to characterize a reaction in either direction. However, the more negative reaction must serve as the electron donor for a spontaneous reaction to occur.
6. The standard cell potential can be calculated by finding the difference between the two half-cell potentials
7. E⁰_cell = E⁰_reduction - E⁰_oxidation
8. E⁰_cell = (0.15 V) - (-0.76 V) = 0.91 V
9. This adds up with our rationale and reaction in part 1, as a cell potential of 0.91 V implies that the reaction is spontaneous/galvanic because it is > 0. The "0" superscript implies cell potential at standard conditions.
10. The Nernst equation can be used to adjust a cell voltage from standard conditions to a different set of thermodynamic conditions.
11. E_cell = E_cell⁰ - (RT/nF)*ln Q
12. E_cell is cell potential in non-standard conditions, E⁰_cell is standard cell potential, R is gas constant, T is temperature, n is # of electrons in reaction, F is Faraday's constant, and Q is the reaction quotient
13. To answer this question, we have all values except T, so we will assume T = 298 K
14. Initial Zinc Concentration
15. E_cell = (0.91 V) - ((8.314 J/mol K)*(298K))/((2 mol electron)*(96485 C/mol electron)* ln ([Mn₂O₃] [OH^-]² [Zn²⁺] / [MnO₂]² [H₂O] [Zn]) = 1.51303 V
16. Because Mn₂O₃, MnO₂, H₂O, and Zn are pure substances in the solid or liquid phase, we can assume their activity (also generally called concentration) is unity of value 1. Insert given values for OH^- and Zn²⁺
17. 10x Zinc Concentration
18. E_cell = (0.91 V) - ((8.314 J/mol K)*(298K))/((2 mol electron)*(96485 C/mol electron)* ln ([Mn₂O₃] [OH^-]² [Zn²⁺] / [MnO₂]² [H₂O] [Zn]) = 1.48347 V
19. Insert given values for OH^- and Zn²⁺ but multiply [Zn²⁺] by 10
20. (1.48347 V)/(1.51303 V) * 100 = 98.046% of the initial voltage remains after increasing zinc concentration 10x.