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Question:
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In the button cells widely used in watches and other devices the following reaction takes place:
\(\mathrm{Zn}(s)+\mathrm{Ag}_2 \mathrm{O}(s)+\mathrm{H}_2 \mathrm{O}(l)\)
\(\longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)+2 \mathrm{OH}^{-}(a q)\)
Determine \(\Delta_r G^{\circ}\) and \(E^{\circ}\) for the reaction.
\(\mathrm{Zn}(s)+\mathrm{Ag}_2 \mathrm{O}(s)+\mathrm{H}_2 \mathrm{O}(l)\)
\(\longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)+2 \mathrm{OH}^{-}(a q)\)
Determine \(\Delta_r G^{\circ}\) and \(E^{\circ}\) for the reaction.
Solution:
1646 Upvotes
Verified Answer
\(\mathrm{Zn}\) is oxidized and \(\mathrm{Ag}_2 \mathrm{O}\) is reduced.
\(\begin{aligned}
E_{\text {cell }}^{\circ} &=E_{\mathrm{Ag}_2 \mathrm{O}, \mathrm{Ag} \text { (reduction) }}^{\circ}-E_{\mathrm{Zn} / \mathrm{Zn}^{2+} \text { (oxidation) }}^{\circ} \\
&=0 \cdot 344+0.76=1 \cdot 104 \mathrm{~V} \\
\Delta \mathrm{G} &=-n F E_{\text {cell }}^{\circ}=-2 \times 96500 \times 1 \cdot 104 \mathrm{~J} \\
&=-2 \cdot 13 \times 10^5 \mathrm{~J} .
\end{aligned}\)
\(\begin{aligned}
E_{\text {cell }}^{\circ} &=E_{\mathrm{Ag}_2 \mathrm{O}, \mathrm{Ag} \text { (reduction) }}^{\circ}-E_{\mathrm{Zn} / \mathrm{Zn}^{2+} \text { (oxidation) }}^{\circ} \\
&=0 \cdot 344+0.76=1 \cdot 104 \mathrm{~V} \\
\Delta \mathrm{G} &=-n F E_{\text {cell }}^{\circ}=-2 \times 96500 \times 1 \cdot 104 \mathrm{~J} \\
&=-2 \cdot 13 \times 10^5 \mathrm{~J} .
\end{aligned}\)
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