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NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

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Question : 23
Total: 32
Calculate the emf of the cell in which the following reaction takes place:
Ni(s)+2Ag+(0.002M) –––––––▶Ni2+(0.160M)+2Ag(s)
Given thatE°cell= 1.05 V
Solution:  
Ni(s)→Ni2+(aq)+2e−
2Ag+(aq)++2e−→2 Ag(s)
–––––––––––––––––––––––––––––––
Ni(s)+2Ag+(aq)→Ni2+(aq)+2Ag(s)
–––––––––––––––––––––––––––––––
E° cell=E°cell−
0.0591
2
‌log‌ ‌
[ Ni2+]
[ Ag+]2

=1.05−
0.0591
2
‌log‌ ‌
0.16
(0.002)2

=1.05−
0.0591
2
‌log‌
0.16
4×10−6

=1.05−
0.0591
2
‌log‌4×104
=1.05−
0.0591
2
(log‌4+log‌104)
=1.05−
0.0591
2
(0.602+4.0000)
=1.05−
0.0591×4.602
2

=1.05–0.1358=0.9142
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