Question

Calculate the emf of the following cell at 250C

Zn/Zn 2+(0.04)//Cd 2+ (0.2) /Cd

Given E0 Zn=-0.76Vand E0 Cd=-0.403V


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Answer 1

Correct option (c) E=+0.36 - (0.059/2)log 2×102

Explanation:

E=E - 0.059/n log [oxdised state]/[reduced state]

cell

n= number of electrons taken part in the reaction is equal to 2

E=[E 2+E

zn/zn cd/cd] -0.059log 0.004/0.2

= [ 0.763 - 0.403] - 0.059/2 log 1/50

=0.36 - 0.059/2 log 2× -2

10

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