Q79 / 180 · NEET 2013 (All India) · Chemistry

A button cell used in watches functions as following $\mathrm{Zn}(s)+\mathrm{Ag}_2 \mathrm{O}(s) & +\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons 2 \mathrm{Ag}(s) + \mathrm{Zn}^{2+}(\alpha q)+2 \mathrm{OH}^{-}(\alpha q)$ 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

  • Correct answer
    A
    $1.10 \mathrm{~V}$
  • B
    $0.42 \mathrm{~V}$
  • C
    $0.84 \mathrm{~V}$
  • D
    $1.34 \mathrm{~V}$

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 $\begin{aligned} & \mathrm{Ag}_2 \mathrm{O}(s)+ \mathrm{H}_2 \mathrm{O}(l)+2 e^{-} \longrightarrow 2 \mathrm{Ag}(s)+2 \mathrm{OH} \\ & E^{-}(a q) ; E^{\circ}=0.34 \mathrm{~V} \\ &= E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ &= 0.34-(-0.76)=+1.10 \mathrm{~V} \end{aligned}$

Hence, the correct answer is option: 'A'.

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