Q.

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

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

The value of x is ________. (Round off to the nearest integer)   

[Give: EZn2+/Zn=0.76V;EAg+/Ag=+0.80V;2.303RTF=0.059]

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answer is 147.

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Detailed Solution

Considering the given cell-representation:

Oxidation at anode:

Zn(s)→Zn2+(aq)+2e
Reduction at cathode: 

Ag+(aq)+e→Ag(s)

 Ecell o=EAg+/AgoEZn2+/ZnoEcell o=0.80(0.76) V=+1.56VZn(s)+2Ag+(aq)Zn(aq)+2Ag(s)(n=2)QR=Zn2+Ag+2Ecell =Ecell o0.059nlog QR=1.560.0592log0.10.01=1.560.08865 =1.47135 V=147.135×102 V147×102 V

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