KVPY SX SB Exam 2013 Question Paper

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Question : 105 of 120

Marks: +1, -0

The standard electrode potential of

Zn^{2+}\/Zn

is - 0.76 V and that of

Cu^{2+}\/Cu

is 0.34 V. The emf (V) and the free energy change

(\kJ \mol^{-1})

, respectively, for a Daniel cell will be

-0.42 and 81
1.1 and -213
-1.1 and 213
0.42 and -81

Solution:

E°_\Cell=E°_{\Cu^{2+}|\Cu}+E°_{\Zn|\Zn^{2+}}

E°_{\cell}=0.34+0.76=+1.1\V

[∵\Δ\G=-\nFE°_\cell]

ΔG = - 2 × 96500 × 1.1 = - 213 V

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