(a)(i) Draw and label a simple cell for the electrolytic purification of copper. (ii) Write can equation for the reaction at each electrode in (a)(i) above....

(a)(i) Electrolytic purification of copper

Impure copper is made the anode and pure copper is the cathode. Both are immersed in acidified copper(II) sulfate solution and connected to a direct-current supply. Pure copper is deposited on the cathode, while insoluble impurities settle as anode mud.

(ii) Electrode reactions

Anode (oxidation): \[\mathrm{Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-}\]

Cathode (reduction): \[\mathrm{Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}}\]

(iii) The Daniell cell is an electrochemical (galvanic) cell because it converts chemical energy from a spontaneous redox reaction into electrical energy.

(iv) In the Leclanché cell, \(\mathrm{MnO_2}\) acts as a depolariser. It removes hydrogen formed at the carbon electrode, thereby preventing polarisation and a fall in voltage.

(b)(i)

Balancing oxygen atoms gives \(y=4\), and balancing charge gives \(x=5\):

\[\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}\]

Therefore, \(\boxed{x=5}\) and \(\boxed{y=4}\).

(ii) Let the oxidation state of Mn be \(z\):

\[z+4(-2)=-1\]

\[z=+7\]

Hence, the oxidation state of Mn in \(\mathrm{MnO_4^-}\) is \(\boxed{+7}\).

(c)(i) Factors affecting selective discharge of ions

1. Position of the ions in the electrochemical series.
2. Concentration of the ions in the electrolyte.
3. Nature of the electrode.

(ii) Faraday's second law of electrolysis

When the same quantity of electricity passes through different electrolytes, the masses of substances liberated or deposited are proportional to their chemical equivalent masses.

(iii) Calculation

Since the voltameters are connected in series, the same quantity of electricity passes through both solutions.

For silver:

\[\mathrm{Ag^+ + e^- \rightarrow Ag}\]

\[\text{Moles of Ag deposited}=\frac{0.780}{108}=7.222\times10^{-3}\ \text{mol}\]

One mole of Ag requires one mole of electrons. Therefore:

\[\text{Moles of electrons}=7.222\times10^{-3}\ \text{mol}\]

I. Mass of copper deposited

\[\mathrm{Cu^{2+}+2e^-\rightarrow Cu}\]

\[\text{Moles of Cu}=\frac{7.222\times10^{-3}}{2}=3.611\times10^{-3}\ \text{mol}\]

\[\text{Mass of Cu}=3.611\times10^{-3}\times63.5=0.229\ \text{g}\]

\[\boxed{\text{Mass of copper deposited}=0.229\ \text{g}}\]

II. Quantity of electricity and time of flow

\[Q=nF=(7.222\times10^{-3})(96500)=696.9\ \text{C}\]

\[\boxed{Q\approx697\ \text{C}}\]

Since \(Q=It\):

\[t=\frac{Q}{I}=\frac{696.9}{0.200}=3484.5\ \text{s}\]

\[\boxed{t\approx3.48\times10^3\ \text{s}=58.1\ \text{min}}\]