Question 13 Marks
In the button cell widely used in watches and other devices the following reaction takes place:
$
Zn(s)+Ag_2 O(s)+H_2 O(l) \rightarrow Zn^{2+}(a q)+2 Ag(s)+2 OH^{-}(a q)
$
Determine $\Delta_r G^{(-)}$and $E ^{(-)}$for the reaction
Given $Z n \rightarrow Z n^{2+}+2 e^{-}, E ^0=0.76 V$
Given $Ag \rightarrow Ag ^{+}+2 e^{-}, E ^0=0.344 V$
$
Zn(s)+Ag_2 O(s)+H_2 O(l) \rightarrow Zn^{2+}(a q)+2 Ag(s)+2 OH^{-}(a q)
$
Determine $\Delta_r G^{(-)}$and $E ^{(-)}$for the reaction
Given $Z n \rightarrow Z n^{2+}+2 e^{-}, E ^0=0.76 V$
Given $Ag \rightarrow Ag ^{+}+2 e^{-}, E ^0=0.344 V$
Answer
View full question & answer→Zn is oxidized and $Ag _2 O$ is reduced (as $Ag ^{+}$ions change to Ag )
$
\begin{array}{l}
E_{\text {cell }}^0=E^0\left[A g_2 O / A g\right](r e d)+E^0\left[Z n / Z n^{2+}\right](o x) \\
=0.344+0.76 \\
=1.104 V \\
\Delta_r G^0=-n F E^0 \text { cell }=-2 \times 96500 \times 1.104 J \\
=-2.13 \times 10^5 J
\end{array}
$
$
\begin{array}{l}
E_{\text {cell }}^0=E^0\left[A g_2 O / A g\right](r e d)+E^0\left[Z n / Z n^{2+}\right](o x) \\
=0.344+0.76 \\
=1.104 V \\
\Delta_r G^0=-n F E^0 \text { cell }=-2 \times 96500 \times 1.104 J \\
=-2.13 \times 10^5 J
\end{array}
$




