Given the following half-reactions occurring in the silver-zinc hearing aid battery,calculate the voltage at 25 °C of this battery.
ZnO(s) + H₂O(l) + 2 e^- → Zn(s) + 2 OH^-(aq) E° = -1.260 V
Ag₂O(s) + H₂O(l) + 2 e^- → 2 Ag(s) + 2 OH^-(aq) E° = +0.342 V

Question

Quizplus · Accepted Answer

To calculate the voltage of the battery, we must first identify the anode and cathode. 
In the first half-reaction, Zn is being oxidized into Zn2+ ions and releasing electrons. Therefore, this reaction is the anode:
Zn(s) → Zn2+(aq) + 2e-    E° = -1.260 V
In the second half-reaction, Ag+ ions are being reduced and gaining electrons. Therefore, this reaction is the cathode:
2Ag+(aq) + 2e- → 2Ag(s)     E° = +0.342 V

To calculate the overall voltage of the battery, we can subtract the anode potential from the cathode potential:
E°cell = E°cathode - E°anode
E°cell = (+0.342 V) - (-1.260 V)
E°cell = +1.602 V

Therefore, the best choice is A) +1.602 V.

Given the Following Half-Reactions Occurring in the Silver-Zinc Hearing Aid