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Question

What will be the effect on emf of given cell when the concentration of Ag+ at cathode is decreased to 0.1 M?
Ag(s)|Ag+(aq, 0.001 M)||Ag+(aq, 1 M)|Ag(s)

A
Emf of the cell increased from 0.118 to 0.214 V
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B
Emf of the cell gets reduced from 0.177 to 0.118 V
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C
Emf of the cell gets reduced from 0.214 to 0 V
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D
Emf of the cell increased from 0.117 to 3.31 V
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Solution

The correct option is B Emf of the cell gets reduced from 0.177 to 0.118 V
From cell representation,
[Ag+]=0.001 M Anode
[Ag+]=1 M Cathode

Cell reaction will be,
Ag+(aq, 1 M)Ag+(aq, 0.001 M)

By nernst equation,

Ecell=E0cell0.05911 log [0.001][1]

Ecell=E0cell+0.05911log[1][0.001]

For concentration cells, cathode and anode consist of same metal and its solution​.
Thus,
E0=0

Ecell=0.05911log[1][0.001]

Ecell=0.05911 log 103

Ecell=0.05911×3

Ecell0.177 V

Now, concentration of Ag+ at cathode is reduced to 0.1 M

Cell reaction will be,
Ag+(aq, 0.1 M)Ag+(aq, 0.001 M)

Ecell=E0cell0.05911 log [0.001][0.1]

Ecell=E0cell+0.05911log[0.1][0.001]

E0=0

Ecell=0.05911log[0.1][0.001]

Ecell=0.05911 log 102

Ecell=0.05911×2

Ecell0.118 V

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