Will(cathode) =0.19 Vreduce $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{M}\mathbf{}{\mathbf{SbO}}^{\mathbf{+}}$at pH 2 by the reaction${\mathbf{SbO}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{3}{\mathbf{e}}^{\mathbf{-}}\mathbf{\rightleftharpoons }\mathbf{Sb}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{,}{\mathbf{E}}^{\mathbf{°}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{208}\mathbf{V}$?

When the electric current is too small, the voltage of cell is given as

E=E( cathode )-E( anode)

E( cathode) is electrode's potential which is attached to the negative terminal of the current source.

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

Overpotential:The activation energy of a reaction at an electrode can be overcome by voltage. The required voltage to apply is called overpotential.

Ohmic potential: In an electrochemical cell, the electrical resistance of a solution while current I flows can be overcome by voltage. The required voltage to apply is called ohmic potential.

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

Concentration Polarization: It is the change in concentration of products and reactants at the electrode's surface unlike they are the same in solution.

Given,

The concentration of${\mathrm{SbO}}^{+}$ is 0.1M.

E( cathode$)0.208-\frac{0.05916}{2}\log \frac{1}{0.10}=0.11\mathrm{V}$

In order to reduce ${\mathrm{SbO}}^{+},0.11\mathrm{V}$is required. But the given E (cathode) for reducing is 0.19 V. Thus,${\mathrm{SbO}}^{+}$ reduction does not happen.