Question

The e.m.f. of a Daniell cell at $298\, K$ is $E _{1}$. $Zn / SO _{4}(0.01 M ) \| CuSO _{4}(1.0 M ) / Cu$ When the concentration of $ZnSO _{4}$ is $1.0\, M$ and that of $CuSO _{4}$ is $0.01\, M$, the e.m.f. is changed to $E _{2}$. What is the relationship between $E _{1}$ and $E _{2}$ :

• A. $E_1 < E_2$
• B. $E_1 > E_2$
• C. $E_2 = 0 \neq E_1$
• D. $E_1 = E_2$

Answer 1

Answer: (B)

$E_1 > E_2$

Answer 2

$Zn \left| ZnSO _{4}(0.01\, M ) \| CuSO _{4}(1.0\, M )\right| Cu$
$\therefore E _{1}= E _{ cell }^{\circ}-\frac{2.303\, RT }{2 \times F } \times \log \frac{(0.01)}{1}$

When concentrations are changed

$\therefore E _{2}= E _{ cell }^{\circ}-\frac{2.303 RT }{2 F } \times \log \frac{1}{0.01}$

i.e., $E _{1}> E _{2}$