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Choose the Best Answer · Q3

Q.The button cell used in watches functions as follows: Zn(s)+Ag2O(s)+H2O(l)⇌2Ag(s)+Zn2+(aq)+2OH−(aq)\text{Zn(s)} + \text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightleftharpoons 2\text{Ag(s)} + \text{Zn}^{2+}\text{(aq)} + 2\text{OH}^{-}\text{(aq)} The half-cell potentials are Ag2O(s)+H2O(l)+2e−→2Ag(s)+2OH−(aq)\text{Ag}_2\text{O(s)} + \text{H}_2\text{O(l)} + 2e^{-} \rightarrow 2\text{Ag(s)} + 2\text{OH}^{-}\text{(aq)}, Eo=0.34E^{o} = 0.34 V, and Zn(s)→Zn2+(aq)+2e−\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2e^{-}, Eo=0.76E^{o} = 0.76 V. The cell potential will be

(a) 0.84 V
(b) 1.34 V
(c) 1.10 V
(d) 0.42 V
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✓ Free question

Step 1. Identify the two half-reactions and their roles. The zinc half-reaction, Zn(s)→Zn2+(aq)+2e−\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)}+2e^-, Eo=0.76E^{o}=0.76 V, is an OXIDATION as written (zinc loses electrons), so this electrode is the anode and 0.76 V is already the oxidation potential.

Step 2. The silver-oxide half-reaction, Ag2O(s)+H2O(l)+2e−→2Ag(s)+2OH−(aq)\text{Ag}_2\text{O(s)}+\text{H}_2\text{O(l)}+2e^- \rightarrow 2\text{Ag(s)}+2\text{OH}^-\text{(aq)}, Eo=0.34E^{o}=0.34 V, is a REDUCTION as written (it gains electrons), so this electrode is the cathode and 0.34 V is the reduction potential.

Step 3. Add the oxidation potential of the anode to the reduction potential of the cathode: Ecell=Eox(anode)+Ered(cathode)=0.76+0.34=1.10E_{cell} = E_{ox}(\text{anode}) + E_{red}(\text{cathode}) = 0.76 + 0.34 = 1.10 V.

✓Final answer

The cell potential is 1.10 V, option (c).

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