(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....
Question 1 Report
(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.
(iii) State with reason whether the Daniell cell is an electrolytic cell or an electrochemical cell.
(iv) What is the function of MnO\(_2\) in the Laclanche cell?
(b) Consider the following equation: MnO\(^-_4\) + 8H\(^+\) + xe\(^-\) \(\to\) Mn\(^{2+}\) + yH\(_2\)O. State the
(i) values of x and y;
(ii) oxidation state of Mn in MnO\(^-_4\).
(c)(i) List three factors that affect selective discharge of ions during electrolysis
(ii) State Faraday's second law of electrolysis.
(iii) A voltameter containing silver trioxonitrate(V) solution was connected in series to another voltameter containing copper (II) tetraoxosulphate(VI) solution. When a current ri 0.200 ampere was passed through the solutions, 0.780g of silver was deposited. Calculate the
I. mass of copper that would be deposited in the copper voltameter
II. quantity of electricity used and the time of current flow. [Cu = 63.5 ; Ag = 108; 1F = 96500C]
Answer Details
(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
- Position of the ions in the electrochemical series.
- Concentration of the ions in the electrolyte.
- 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 \(Q=It\):
\[t=\frac{Q}{I}=\frac{696.9}{0.200}=3484.5\ \text{s}\]
\[\boxed{t\approx3.48\times10^3\ \text{s}=58.1\ \text{min}}\]
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