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NCERT Class XII Chemistry
Chapter - Electrochemistry
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Question : 6 of 32
Marks: +1, -0
In the button cells widely used in watches and other devices the following reaction takes place :
Zn(s)+Ag2O(s)+H2O(l)→Zn2+(aq)+2Ag(s)+2OH−(aq)\mathrm{Zn}_{(s)} + \mathrm{Ag_2O}_{(s)} + \mathrm{H_2O}_{(l)} \rightarrow \mathrm{Zn^{2+}}_{(aq)} + 2\mathrm{Ag}_{(s)} + \mathrm{2OH^{-}}_{(aq)}
Determine ΔrG∘\Delta_r G^\circ and E° for the reaction.
Given : EAg2O/Ag∘=0.344 V,EZn2+/Zn∘=−0.76 V,E^\circ_{\mathrm{Ag_2O/Ag}} = 0.344\,\text{V}, E^\circ_{\mathrm{Zn^{2+}}/\mathrm{Zn}} = -0.76\,\text{V},
Solution:  
Zn\mathrm{Zn} is oxidised and Ag2O\mathrm{Ag_2O} is reduced (as Ag+\mathrm{Ag}^{+}ions change into Ag\mathrm{Ag})
Ecell∘=EAg2O/Ag∘−EZn2+/Zn∘E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{Ag_2O/Ag}} - E^\circ_{\mathrm{Zn^{2+}}/\mathrm{Zn}}
=0.344−(−0.76)=1.104 V= 0.344 - (-0.76) = 1.104\,\text{V}
ΔrG∘=−nFEcell∘\Delta_r G^\circ = -n F E^\circ_{\mathrm{cell}}
=−2×96500×1.104= -2 \times 96500 \times 1.104
=−2.13×105 J= -2.13 \times 10^{5}\,\text{J}
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