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A button cell used in watches functions as following: 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+}(aq) + 2\text{OH}^-(aq) If half cell potentials are: Zn2+(aq)+2eZn(s);E=0.76 V\text{Zn}^{2+}(aq) + 2e^- \rightarrow \text{Zn}(s); E^\circ = -0.76 \text{ V} Ag2O(s)+H2O(l)+2e2Ag(s)+2OH(aq);E=0.34 V\text{Ag}_2\text{O}(s) + \text{H}_2\text{O}(l) + 2e^- \rightarrow 2\text{Ag}(s) + 2\text{OH}^-(aq); E^\circ = 0.34 \text{ V} The cell potential will be:

A

0.84 V

B

1.34 V

C

1.10 V

D

0.42 V

Step-by-Step Solution

In the given button cell reaction, Zinc (Zn\text{Zn}) is oxidised to Zn2+\text{Zn}^{2+} and silver oxide (Ag2O\text{Ag}_2\text{O}) is reduced to silver (Ag\text{Ag}). Therefore, the zinc electrode acts as the anode and the silver electrode acts as the cathode. The standard cell potential (EcellE^\circ_{\text{cell}}) is calculated using the formula: Ecell=EcathodeEanodeE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} Given the standard reduction potentials: Ecathode=0.34 VE^\circ_{\text{cathode}} = 0.34 \text{ V} Eanode=0.76 VE^\circ_{\text{anode}} = -0.76 \text{ V} Substituting these values into the formula: Ecell=0.34 V(0.76 V)=0.34+0.76=1.10 VE^\circ_{\text{cell}} = 0.34 \text{ V} - (-0.76 \text{ V}) = 0.34 + 0.76 = 1.10 \text{ V} The standard cell potential is 1.10 V1.10 \text{ V}.

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