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Question 50 of 52

Q.(i) In button cell, widely used in watches, the following reaction takes place: Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH⁻(aq). Determine E°cell and ΔG° for the cell. (Given E°(Ag+/Ag) = +0.80 V and E°(Zn2+/Zn) = -0.76 V.) [2]

(ii) What advantage do the fuel cells have over primary and secondary batteries? [1]
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2025Subjective· 3mImportance★★★★★
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Ecell∘=Ecathode∘−Eanode∘=1.56 VE^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=1.56\ V; ΔG∘=−nFEcell∘≈−301.1 kJ/mol\Delta G^{\circ}=-nFE^{\circ}_{cell}\approx-301.1\ kJ/mol.

(i) In the button cell reaction Zn(s)+Ag2O(s)+H2O(l)→Zn2+(aq)+2Ag(s)+2OH−(aq)Zn(s) + Ag_2O(s) + H_2O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^-(aq), zinc is oxidised (anode: Zn→Zn2++2e−Zn \rightarrow Zn^{2+}+2e^-) and silver(I) is reduced to metallic silver (cathode). This is a 2-electron transfer reaction (n=2n=2).

Ecell∘=Ecathode∘−Eanode∘=E∘(Ag+/Ag)−E∘(Zn2+/Zn)=0.80 V−(−0.76 V)=1.56 VE^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = E^{\circ}(Ag^+/Ag) - E^{\circ}(Zn^{2+}/Zn) = 0.80\ V - (-0.76\ V) = 1.56\ V

Using ΔG∘=−nFEcell∘\Delta G^{\circ} = -nFE^{\circ}_{cell} with F=96500 C mol−1F = 96500\ C\,mol^{-1}:

ΔG∘=−2×96500×1.56≈−301,080 J/mol≈−301.1 kJ/mol\Delta G^{\circ} = -2 \times 96500 \times 1.56 \approx -301{,}080\ J/mol \approx -301.1\ kJ/mol

The large negative ΔG∘\Delta G^{\circ} confirms the cell reaction is strongly spontaneous, consistent with the button cell working reliably as a power source.

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