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Q.

Emf of the following cell at 298 K in V is x×10−2Zn|Zn2+0.1M||Ag+0.01M|AgThe value of x is ________. (Round off to the nearest integer)   [Give: EZn2+/Znθ=−0.76 V; EAg+/Agθ=+0.80 V; 2.303RTF=0.059]

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answer is 153.

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Detailed Solution

ZnZn2+(0.1M)∥Ag+(0.01M)AgEcell o=ECo−EAo=0.80−(−0.76)=+1.56V Cell lxn:Zn(s)+2Ag+(aq)→Zn(aq)+2Ag(s)(n=2)QR=Zn2+Ag+2Ecell =Ecall o−0.059nlog⁡QR=1.56−0.0592log⁡0.10.01=1.56−0.0295=1.5305V=153.05×10−2V
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