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Question

emf of the following cell at 298 K in V is x ×102

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

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

Given: E0Zn2+/Zn=0.76V;E0Ag+/Ag=+0.80V;2.303RTF=0.059


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Solution

Zn(s)|Zn+2(0.1M)||Ag+(0.01M)|Ag(s)
Zn(s)+2Ag+2Ag(s)+Zn+2

E0=0.80+0.76=1.56;
Q=Zn2+(Ag+)2
E=E00.059nlog(Q)

E=1.560.0592log[0.1(0.01)2]

E=1.560.0592log[(10)3]

E=1.4715=147.15×102volt=x×102

x=147.15147


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