Electrochemistry

Key idea According to Nernst equation,

$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{2.303}{n F} R T \log \frac{\left[M_p\right]^{n+}}{\left[M_r\right]^{n+}}$

where, $\left[M_p\right]^{n+}$ and $\left[M_r\right]^{n+}$ are the concentrations of products and reactants respectively.

For the given cell, emf can be calculated as follows:

$E_{\mathrm{emf}}=E_{\mathrm{cell}}^{\circ}-\frac{2.303 R T}{n F} \log \frac{\left[M_p\right]_{(a q)}^{\mathrm{n}+}}{\left[M_r\right]_{(a q)}^{\mathrm{n}+}}$

For the reaction at $25^{\circ} \mathrm{C}$

$\begin{aligned} E_{\text {emf }} & =E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[M_p^{n-1}\right]}{\left[M_r^{n+}\right]} \\ \therefore \quad & E_{\text {cell }}^{\circ}=4 \text { Volt (Given) and } n=2 \\ & E_{\text {emf }}=4-\frac{0.059}{2} \log \frac{10^{-4}}{10^{-2}} \\ & E_{\text {emf }}=4-0.03 \log 10^{-2} \\ E_{\text {emf }} & =4+0.06=4.06 \mathrm{~V} \end{aligned}$

Hence, option (b) is the correct.

Given,

$\begin{aligned} \Delta G_f^{\circ}\left(\mathrm{Ag}_2 \mathrm{O}\right) & =-11.21 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta G_f^{\circ}(\mathrm{ZnO}) & =-318.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

As we know that,

$\Delta G^{\circ}=\Delta G_f^{\circ}(\mathrm{ZnO})-\Delta G_f^{\circ}\left(\mathrm{Ag}_2 \mathrm{O}\right)$

Putting the values in the above given equation, we get

$\begin{array}{lc} \therefore & \Delta G^{\circ}=(-318.30+11.21) \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & =-307.09 \mathrm{~kJ}=-307.09 \times 10^3 \mathrm{~J} \\ \because & \Delta G^{\circ}=-n F E_{\text {cell }}^{\circ} \\ \therefore & -307.09 \times 10^3=-2 \times 96500 \times E_{\text {cell }}^{\circ} \\ \text { or } & E_{\text {cell }}^{\circ}=1.591 \mathrm{~V} \end{array}$

Given,

$\begin{aligned} \lambda_{\mathrm{m}}^{\circ}(\mathrm{NaOH}) & =248 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1} \\ \lambda_{\mathrm{m}}^{\circ}(\mathrm{NaCl}) & =126 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1} \\ \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{BaCl}_2\right) & =280 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1} \end{aligned}$

The reaction for the formation of $\mathrm{Ba}(\mathrm{OH})_2$ can be written as

$\begin{aligned} & \mathrm{BaCl}_2+2 \mathrm{NaOH} \longrightarrow \mathrm{Ba}(\mathrm{OH})_2+2 \mathrm{NaCl} \\ & \because \lambda_{\mathrm{m}}^{\circ} \mathrm{Ba}(\mathrm{OH})_2=\lambda_{\mathrm{m}}^{\circ} \mathrm{BaCl}_2+2 \lambda_{\mathrm{m}}^{\circ} \mathrm{NaOH}-2 \lambda_{\mathrm{m}}^{\circ} \mathrm{NaCl} \\ & \therefore \lambda_{\mathrm{m}}^{\circ} \mathrm{Ba}(\mathrm{OH})_2=280 \times 10^{-4}+2 \times 248 \times 10^{-4} \\ &-2 \times 126 \times 10^{-4} \\ &=(280+496-252) \times 10^{-4} \\ &=524 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1} \end{aligned}$

Both assertion and reason are correct and reason is the correct explanation of assertion. On applying external voltage greater than 1.1V in a Daniell cell, the current flows in the reverse direction i.e. from Zn to Cu (anode to cathode). The electrons flow from copper to zinc. Zn is deposited at Zn electrode and Cu dissolves at Cu electrode. The reaction is

$\mathrm{Zn}^{2+}+\mathrm{Cu} \longrightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}$