(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 |
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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. |
(ii)
(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