Question

In the electrochemical cell
$Zn\left|\right.ZnSO_{4}\left(\right.0.01M\left.\right)\left|\right|CuSO_{4}\left(\right.1.0M\left.\right)\left|\right.Cu,$ the emf of this Daniel cell is $E_{1}$ . When the concentration of $ZnSO_{4}$ is changed to 1.0M and that of $CuSO_{4}$ changed to 0.01M, the emf changes to $E_{2}$ . From the followings, which one is the relationship between $E_{1}$ and $E_{2}$ ?
(Given, $\frac{R T}{F}=0.059$ )

• 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)

$\text{E} \, \text{=E}^{\text{o}}-\frac{0.059}{2}log_{10}\frac{\left[\text{Zn}\right]^{2 +}}{\left[\text{Cu}\right]^{2 +}}$
$\left(\text{E}\right)_{1} = \text{E} ^\circ - \frac{0.059}{2} \left(\text{log}\right)_{10} \frac{\left(0.01\right)}{\left(\right. 1 \left.\right)^{2}}$
$\left(\text{E}\right)_{2} = \text{E} ^\circ - \frac{0.059}{2} \left(\text{log}\right)_{10} \frac{\left(1\right)}{\left(\right. 0.01 \left.\right)^{2}}$
Hence $E_{1}>E_{2}$