Question

The emf of a Daniel cell at 298 K is $E_{1}.\qquad$The cell is
$Zn | ZnSO_{4} \left(0.01M\right)|| CuSO_{4} \left(1M\right) | CU$
When the concentration of $ZnSO_{4}$ is changed to 1M and that of $CuSO_{4}$ to 0.01M, the emf changes to E2.The relationship between $E_{1} and E_{2}$ will be

• A. $E_{1}-E_{2} =0$
• B. $E_{1} < E_{2}$
• C. $E_{1}>E_{2}$
• D. $E_{1}=10^{2} E_{2}$

Answer 1

Answer: (C)

$Zn+CuSO_{4} {\rightleftharpoons} ZnSO_{4} +cu$
n = 2
$Q=\frac{Zn^{2+}}{cu^{2+}}$
$E_{1}-E_{2} =-\frac{0.059}{2} \left(log\frac{0.01}{1}log\frac{1}{0.01}\right) =-\frac{0.059}{2}\left(log\frac{1}{100}-log100\right)=\frac{0.059}{2} \left(-log100 -log 100\right)$$=+\frac{0.059}{2}\times2$
$so, E_{1} >E_{2}$