(a)(i) Define oxidation in terms of electron transfer. (ii) Write balanced equations for the half...
(a)(i) Define oxidation in terms of electron transfer.
(ii) Write balanced equations for the half reactions for the following changes in acidic solution: Mn0\(^-_4\) + Fe\(^{2+}\) —> Mn\(^{2+}\) + Fe\(^{3+}\)
(b)(i) Distinguish between an electrolytic celI and an electrochemical cell.
(ii) Sketch a cell for the electrolysis of molten magnesium chloride. Lable the anode and the cathode and indicate the direction of electron flow. Give the electrode reactions.
(iii) Give one reason why a platinum anode is not suitable for the eloctrolysis in (b)(i) above.
(c) Calculate the mass of lead that would be deposited from a solution of lead (II) trioxonitrate by the same quantity of electrically depositing 1.35g of copper. (Cu = 63.5, Pb = 207)
Explanation
(a)(i) Oxidation is the process of electron loss. E.g \(Fe_{(g)} \to Fe^{2+}_{(aq)} + 2e^-\)
(ii) \(Fe^{2+}_{(aq)} + e^-\) Oxidation half reaction
\(MnO^- _{4(aq)} + 8H^+_{(aq)} + 8e^- \to Mn^{2+}_{(aq)} + 4H_2O\)
Reduction half rxn.
(b)(i) Difference between electrolytic cell and electrochemical cell.
Electrolytic cell | Electrochemical cell |
Electrica energy is use to ring about chemical energy. Cathode electrode is negatively charged. | Chemical energy is used to bring about electrical energy. The anode is the negatively charged electrode. |
(iii) The platinum anode is attced by chlorine gas.
(c) If 63.5g Cu is equivalent to \(\frac{207}{63.5}\)
1.35g Cu will be equivalent to \(\frac{207}{63.5} \times \frac{1.35}{1}\) = 4.4g Pb