A button cell used in watches functions as following \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_2…

A button cell used in watches functions as following \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_2 \mathrm{O}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\ell) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+ \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})\)
If half cell potentials are
$\begin{aligned}
\mathrm{Zn}^{2+}(a q)+2 e^{-} & \rightarrow \mathrm{Zn}(s) ; E^{\circ}=-0.76 \mathrm{~V} \\
\mathrm{Ag}_2 \mathrm{O}(s)+\mathrm{H}_2 \mathrm{O}(l) & +2 e^{-} \\
& \rightarrow 2 \mathrm{Ag}(s)+2 \mathrm{OH}^{-}(a q), \\
& E^{\circ}=0.34 \mathrm{~V}
\end{aligned}$ The cell potential will be
  1. $1.10 \mathrm{~V}$
  2. $0.42 \mathrm{~V}$
  3. $0.84 \mathrm{~V}$
  4. $1.34 \mathrm{~V}$

Solution

Anode is always the site of oxidation thus anode half cell is
$\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}(\mathrm{s}) ; \quad E^{\circ}=-0.76 \mathrm{~V}$
Cathode half cell is
$\begin{aligned}
& \mathrm{Ag}_2 \mathrm{O}(s)+ \mathrm{H}_2 \mathrm{O}(l)+2 e^{-} \longrightarrow 2 \mathrm{Ag}(s)+2 \mathrm{OH} \\
& E^{-}(a q) ; E^{\circ}=0.34 \mathrm{~V} \\
&= E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\
&= 0.34-(-0.76)=+1.10 \mathrm{~V}
\end{aligned}$

Asked in: NEET 2013 (All India)

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