Question

A concentration cell consists of two $\mathbf{Sn}\mathbf{/}{\mathbf{Sn}}^{\mathbf{2}\mathbf{+}}$half-cells. The electrolyte in compartment A is 0.13 M $\mathbf{Sn}{\left({\mathbf{NO}}_{\mathbf{3}}\right)}_{\mathbf{2}}$. The electrolyte in B is 0.87M $\mathbf{Sn}{\left({\mathbf{NO}}_{\mathbf{3}}\right)}_{\mathbf{2}}$. Which half-cell houses the cathode? What is the voltage of the cell?

Short answer

$E_{cell}=0.024V$.

Step-by-step solution

Concentration cell

A concentration cell is a cell in which the potential difference is created due to the difference in the concentration of the same chemical species in both the half cells. Since, same chemical species are present in both the half-cells, the standard cell potential for a concentration cell is zero. The standard cell potential for concentration cells is zero because the standard electrode potential for both the half-cells are equal.

To find the voltage of the cell

In a concentration cell, the cathode is the electrode with more concentrated electrolyte solution. Half-cell B contains 0.87M of electrolyte, while half-cell A only has 0.13M. Half-cell B is the cathode.

$$\begin{gathered} \text{Cathode} \text{reaction} {\text{Sn}}^{\text{+ 2}}\left(\text{0.87} \text{M}\right){\text{+ 2e}}^{\text{-}}\to \text{Sn}\left(\text{s}\right) \\ \text{Anode} \text{reaction} \text{Sn}\left(\text{s}\right)\to {\text{Sn}}^{\text{+ 2}}\left(\text{0.13} \text{M}\right){\text{+ 2e}}^{\text{-}} \\ \text{Overall} \text{reaction} {\text{Sn}}^{\text{+ 2}}\left(\text{0.87} \text{M}\right) \to {\text{Sn}}^{\text{+ 2}}\left(\text{0.13} \text{M}\right) \end{gathered}$$

To compute for the voltage of the cell, the concentration of these half-cells must be used in the equation

$$\begin{gathered} E_{cell}={E^0}_{cell}-\frac{\text{RT}}{\text{nF}}\text{ln}\frac{\text{[ ions in anode compartment ]}}{\text{[ions in cathode compartment]}} \\ {E}_{cell}={E^0}_{cell}-\frac{0.0592V}{n}\log \frac{\left[S{n}^{2+}\right]}{\left[S{n}^{2+}\right]} \end{gathered}$$

$$\begin{gathered} E_{cell}=0-\frac{0.592}{2}\log \frac{0.13}{0.87} \\ {E}_{cell}=0.024V \end{gathered}$$

Hence$E_{cell}=0.024V$.