Question

If \[\mathrm{E^o\left(Cd_{(aq)}^{+2}|Cd_{(s)}\right)=-0.40~V}\]. What is potential for \[\mathrm{Cd}_{(s)}\xrightarrow{}\mathrm{Cd}_{(aq)}^{+2}\left(0.01\mathrm{M}\right)+2\mathrm{e}^{-}\] at 298 K?

Options

• +0.4592 V

• -0.4592 V

• +0.3408 V

• -0.3408 V

Answer 1

+0.4592 V

Explanation:

Given \[\mathrm{E}_{\mathrm{red}}^0\] value is standard reduction potential.

∴  \[\mathrm{E}^0\] for oxidation potential = +0.4 V
For the given reaction, \[\mathrm{Cd}_{(\mathrm{s})}\longrightarrow\mathrm{Cd}_{(\mathrm{aq})}^{+2}+2\mathrm{e}^{-}\]
Using Nernst equation,
\[\mathrm{E}_{\mathrm{cell}}=\mathrm{E}^{\mathrm{o}}-\frac{0.0592}{\mathrm{n}}\log_{10}\frac{[\mathrm{Cd}^{+2}]}{1}\]
\[=0.4-\frac{0.0592}{2}\log_{10}(0.01)\]
\[=0.4V+0.0592V\]
\[\therefore\quad\mathrm{E}_{\mathrm{cell}}=0.4592\mathrm{~V}\]