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Question

The emf of the following cells are:
Ag|Ag+(1M) || Cu2+(1M)|Cu   Ecell=0.46V and
Zn|Zn2+(1M) || Cu2+(1M)|Cu   Ecell=+1.10V.
Then the emf of the cell Zn|Zn2+(1M) || Ag+(1M)|Ag is:
  1. 0.64V
  2. 1.1V
  3. 1.56V
  4. – 0 .64V


Solution

The correct option is C 1.56V
Ag|Ag+(1M) || Cu2+(1M)Cu    Ecell=0.46V
E0Cu2+/CuE0Ag+/Ag=0.46V
Zn|Zn2+(1M) || Cu2+(1M)|Cu   Ecell=+1.10V
E0Cu2+/CuE0Zn2+/Zn=1.10V
Zn|Zn2+(1M) || Ag+(1M)|Ag
E0Ag+/AgE0Zn2+/Zn=(E0Cu2+/CuE0Zn2+/Zn)(E0Cu2+/CuE0Ag+/Ag)=1.1(0.46)=1.56V

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