Question

Zinc/silver oxide cell is used in hearing acids and electric watches. The following reactions takes place $ Zn\xrightarrow{{}}Z{{n}^{2+}}+2{{e}^{-}};{{E}^{o}}=0.76V $ $ A{{g}_{2}}O+{{H}_{2}}O+2{{e}^{-}}\xrightarrow{{}}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=-nFE_{cell}^{o}=-2\times 96500\times 1.104J $ $ =-2.13\times {{10}^{5}}J $