(a) At what cathode potential will Sb(s)deposition commence from $\mathbf{0}\mathbf{.}\mathbf{010}{\mathbf{MSbO}}^{\mathbf{+}}$solution at pH 0.00? Express this potential versus S.H.E. and versus$\mathbf{Ag}\mathbf{|}\mathbf{AgCI}$. (Disregard overpotential, about which you have no information.)

${\mathbf{SbO}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{3}{\mathbf{e}}^{\mathbf{-}}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{Sb}\left(s\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{}\mathbf{}\mathbf{E}\mathbf{°}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{208}\mathbf{V}$

(b) What percentage of$\mathbf{0}\mathbf{.}\mathbf{10}{\mathbf{MCu}}^{\mathbf{2}\mathbf{+}}$could be reduced electrolytically to Cu(s)before$\mathbf{0}\mathbf{.}\mathbf{010}{\mathbf{MSbO}}^{\mathbf{+}}$in the same solution begins to be reduced at pH 0.00?

The voltage of a cell when the electric current is too small,

$\mathbf{E}\mathbf{=}\mathbf{E}\left(\mathrm{cathode}\right)\mathbf{-}\mathbf{E}\left(\mathrm{anode}\right)$

E is the electrode's potential that is connected to the current source's negative terminal.

The electrode's potential is E(anode), which is connected to the positive terminal of the current source.

Overpotential: Voltage can override the activation energy of a process at an electrode, resulting in overpotential. Overpotential is the needed voltage to apply.

Ohmic potential: In an electrochemical cell, voltage can overcome the electrical resistance of a solution while currentflows. The ohmic potential is the voltage that must be applied.

${\mathbf{E}}_{\mathbf{ohmic}}\mathbf{=}\mathbf{IR}$

Concentration Polarization: Polarization is defined as a change in product and reactant concentrations at the electrode's surface, although they are the same in solution.

a)

The value of E (cathode)

$$\begin{gathered} =0.208-\frac{0.05916}{2}\log \frac{1}{{\left[{\mathrm{SbO}}^{+}|{\mathrm{HH}}^{+}\right]}^2} \\ =0.208-\frac{0.05916}{2}\log \frac{1}{\left(0.010\right){\left(1.0\right)}^2} \\ =0.169\mathrm{V} \end{gathered}$$

Now,

b)

At 0.169V , the concentration of copper ion equilibrium with copper is calculated as follows${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}\left(\mathrm{s}\right)\mathrm{E}°=0.339$

$$\begin{gathered} \mathrm{E}\left(\mathrm{cathode}\right)=0.339-\frac{0.05916}{2}\log \frac{1}{\left|{\mathrm{Cu}}^{2+}\right|} \\ 0.169=0.339-\frac{0.05916}{2}\log \frac{1}{\left|{\mathrm{Cu}}^{2+}\right|} \\ {\mathrm{Cu}}^{2+1}=1.8\times {10}^{-6}\mathrm{M} \end{gathered}$$

The percentage of copper ions that have not been reduced is calculated as

$$\begin{gathered} =\frac{1.8\times {10}^6}{0.10}\times 100 \\ =1.8\times {10}^{-3\%} \end{gathered}$$

The percentage of reduced copper ion is $99.998\%$.