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Question

Calculate the emf of the following cell at 298 K.
Zn(s)Zn2+(aq, 0.1 M)Ag+(aq, 0.01 M) Ag(s)
Given:
E0Zn2+/Zn = 0.76 V
E0Ag+/Ag =+0.80 V

A
Ecell=1.472 V
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B
Ecell=1.472 V
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C
Ecell=+1.732 V
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D
Ecell=1.732 V
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Solution

The correct option is A Ecell=1.472 V
From the given cell representation,
Ag+/Ag couple act as cathode
Zn2+/Zn couple act as anode

E0cell=E0cathodeE0anode
E0cell=0.80 (0.76)
E0cell=1.56 V
The given cell reaction is,
Zn(s)+2Ag+(aq)Zn2+(aq)+2Ag
n=2
By Nernst equation,
Ecell = Eocell 0.0591nlog[Zn2+][Ag+]2

Ecell = Eocell + 0.0591nlog[Ag+]2[Zn2+]

Ecell= 1.56 + 0.05912log[0.01]2[0.1]

Ecell= 1.56 + 0.05912 log (1 ×103)

Ecell= 1.56 0.05912×3
Ecell= 1.56 0.088 = 1.472 V

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