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Q.

The e.m.f. of the following Daniell cell at 298 K is E₁; Zn /ZnSO₄(0.01M)//CuSO₄ (1.0M)/Cu When the concentration of ZnSO₄ is 1. 0 M and that of CuSO₄is 0.01 M, the e.m.f. changed to E₂. What is the relationship between E₁ and E₂ ?

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a

$E_{1} = E_{2}$

b

$E_{2} = 0 \neq E_{1}$

c

$E_{1} > E_{2}$

d

$E_{1} < E_{2}$

answer is A.

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Detailed Solution

$E = [E^{\circ} - \frac{0.0591}{2} \log \frac{[{Zn}^{+ 2}]}{[{Cu}^{+ 2}]}]$

$E_{1} = E^{0} + 0.06 E_{2} = E^{0} - 0.06$

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The e.m.f. of the following Daniell cell at 298 K is E1; Zn /ZnSO4(0.01M)//CuSO4 (1.0M)/Cu When the concentration of ZnSO4 is 1. 0 M and that of CuSO4 is 0.01 M, the e.m.f. changed to E2. What is the relationship between E1 and E2 ?