Question

Zinc/silver oxide cell is used in hearing acids and electric watches. The following reactions takes place $Zn\to Z{n}^{2+}+2e^{-};E^o=0.76V$ $A{g}_{2}O+H_{2}O+2e^{-}\to 2Ag+2O{H}^{-};$ $E^o=0.344V$ $\Delta G$ in joules will be

• A. $-2.13\times {10}^{5}J$
• B. $+2.13\times {10}^{5}J$
• C. $-4.67\times {10}^{7}J$
• D. $+4.67\times {10}^{7}J$

Answer 1

Answer: (A)

Zinc is oxidised and $A{g}_{2}O$ is reduced. $E_{cell}^o=E_{A{g}_{2}O/Ag}^o(red)+E_{Zn/Z{n}^{2+}}^o(OX)$ $=0.344+0.76=1.104V$ $\Delta H=-nF{E}_{cell}^o=-2\times 96500\times 1.104J$ $=-2.13\times {10}^{5}J$