Question

A solution of Sn²⁺is to be electrolyzed to reduce the Sn²⁺to Sn(s). Calculate the cathode potential (versus S.H.E.) needed to reduce$\left[{\mathrm{Sn}}^{2+}\right]$to$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{M}$if no concentration polarization occurs. What would be the potential versus S.C.E. instead of S.H.E? Would the potential be more positive or more negative if concentration polarization occurred?

Short answer

The potential of cathode versus S.H.E is $\mathrm{E}\left(\mathrm{cathode}\right)=-0.378\mathrm{V}$

The potential of cathode versus S.C.E. is $\mathrm{E}=-0.619\mathrm{V}$

Step-by-step solution

Electrogravimetric Analysis

• Separation along with the quantification of metals are mainly observed in electrogravimetric analysis.
• Electrolysis of the analyte solution is done during this procedure.

Determine the lead lactate

The reaction of cathode

${\mathrm{Sn}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Sn}$

The standard reduction potential is $\mathrm{E}°=-0.141\mathrm{V}$

The potential of cathode versus S.H.E is

$$\begin{gathered} \mathrm{E}\left(\mathrm{cathode}\right)=\mathrm{E}°-\frac{0.05916}{2}·\log \left(\frac{1}{\left[{\mathrm{Sn}}^{2+}\right]}\right) \\ \mathrm{E}\left(\mathrm{cathode}\right)=-0.141\mathrm{V}-\frac{0.05916}{2}·\log \left(\frac{1}{1\times {10}^{-8}\mathrm{M}}\right) \\ \mathrm{E}\left(\mathrm{cathode}\right)=-0.378\mathrm{V} \end{gathered}$$

The potential of cathode versus S.C.E. is

$$\begin{gathered} \mathrm{E}=\mathrm{E}\left(\mathrm{cathode}\right)-\mathrm{E}\left(\mathrm{S}.\mathrm{C}.\mathrm{E}.\right) \\ \mathrm{E}=-0.378\mathrm{V}-0.241\mathrm{V} \\ \mathrm{E}=-0.619\mathrm{V} \end{gathered}$$

The potential would be more negative if concentration polarization occurred.

The potential of cathode versus S.H.E is$\mathrm{E}\left(\mathrm{cathode}\right)=-0.378\mathrm{V}$

The potential of cathode versus S.C.E. is $\mathrm{E}=-0.619\mathrm{V}$