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Question

E.m.f of the following cell at 298K in V is x×10-2.

Zn|Zn2+(0.1M)||Ag+(0.01M)|Ag

The value of x is _________. (Rounded off to the nearest integer)

Given:EZn2+/Zn°=-0.76;EAg+/Ag°=+0.80V;2.303RTF=0.059


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Solution

Step1:CalculationofE°cellE°cell=E°Ag/Ag+-E°Zn2+/Zn=0.80-(-0.76)=1.56VStep2:CalculationofEmfofcelli.e.EcellUsingNernstEquation,Ecell=E°cell-0.0592log[Zn2+][Ag+]2=1.56-0.0592log[0.1][0.01]2{Substitutingthegivenvalues}=1.56-0.0592log3=1.56-0.0885=1.4715=147.15×10-2147×10-2=x×10-2Hence,thevalueofxis147.15V147V.


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