In $\mathbf{1}\mathbf{MN}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}\mathbf{/}\mathbf{1}\mathbf{M}\mathbf{}{\mathbf{NH}}_{\mathbf{4}}\mathbf{}\mathbf{CI}$solution,${\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}$is reduced to${\mathbf{Cu}}^{\mathbf{+}}$near -0.3V(versus S.C.E.), and${\mathbf{Cu}}^{\mathbf{+}}$is reduced to$\mathbf{Cu}\left(\mathrm{inHg}\right)$near 0.6V.

(a) Sketch a qualitative sampled current polarogram for a solution of${\mathbf{Cu}}^{\mathbf{+}}$.

(b) Sketch a polarogram for a solution of${\mathbf{Cu}}^{\mathbf{2}\mathbf{+}}$.

(c) Suppose that Pt, instead of Hg, were used as the working electrode. Which, if any, reduction potential would you expect to change?

The correlation between current and voltage in an electrochemical reaction is noted as voltammetry

Polarogram is a graph representing current vs voltage.

(a)

The reaction is:

$C{u}^{+}\to Cu\left(Hg\right)$The reduction of copper ion leads to copper-amalgam. The graph between current and voltage of reduction is given above.

(b)

The reactions are:

$$\begin{gathered} C{u}^{2+}\to C{u}^{+} \\ C{u}^{+}\to Cu\left(Hg\right) \end{gathered}$$

The reduction of $C{u}^{2+}$forms $C{u}^{+}$and further reduction with increase in voltage leads to copper-amalgam. The graph between current and voltage of reduction is given above.

(c) The reduction potential for the reduction reaction of copper ion to copper will obviously change because the working electrode is changed to platinum. If Pt, instead of , were used as the working electrode, the reduction potential that would we expect to change is$C{u}^{+}\to Cu\left(Hg\right)$ .