(a)(i) Explain briefly each of the following terms: I. anode; II. cathode. (ii) Sodium and aluminium are extracted by the electrolysis of molten sodium chlo...
Question 1 Report
(a)(i) Explain briefly each of the following terms: I. anode; II. cathode.
(ii) Sodium and aluminium are extracted by the electrolysis of molten sodium chloride and alumina respectively. Write balanced equations for the reactions at the anode and cathode during the extraction of: I. sodium; II. aluminium.
(iii) Explain briefly why extraction of aluminium is considered environmentally friendly while that of sodium is not.
(b) Consider the reaction represented by the following equation:
K\(_2\)Cr\(_2\)O\(_7\) + HC1--> KCI + CrCl\(_3\) + H\(_2\)O + Cl\(_2\)
(i) Explain briefly why this reaction is redox.
(ii) Write balanced half equations for the reaction.
(iii) Write the over-all balanced reaction equation.
(c) During the electrolysis of molten Al\(_2\)O\(_3\), a current of 6A was passed through the electrolyte for 1 hr. 30 mins. Calculate the mass of aluminium deposited at the cathode.
Answer Details
(a)(i) I. Anode: the electrode at which oxidation (loss of electrons) occurs; in electrolysis it is the positive electrode, where anions are discharged. II. Cathode: the electrode at which reduction (gain of electrons) occurs; in electrolysis it is the negative electrode, where cations are discharged.
(ii) Extraction equations
I. Sodium (molten NaCl):
Cathode: Na+ + e- → Na
Anode: 2Cl- → Cl2 + 2e-
II. Aluminium (molten Al2O3):
Cathode: Al3+ + 3e- → Al
Anode: 2O2- → O2 + 4e-
(iii) In the extraction of aluminium the gas discharged at the anode is oxygen, which is harmless. In the extraction of sodium the gas discharged at the anode is chlorine, which is poisonous and pollutes the environment. Hence the sodium process releases a toxic gas whereas the aluminium process does not.
(b)(i) The reaction is redox because chromium is reduced (its oxidation number falls from +6 in Cr2O72- to +3 in CrCl3) while chlorine is oxidized (from -1 in HCl to 0 in Cl2); oxidation and reduction occur together.
(ii) Half equations
Reduction: Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
Oxidation: 2Cl- → Cl2 + 2e-
(iii) Overall:
K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 7H2O + 3Cl2
(c) Time = 1 h 30 min = 5400 s, current = 6 A, Al3+ + 3e- → Al.
\[ Q=6\times5400=32400\ \text{C} \] \[ n(e^-)=\frac{32400}{96500}=0.3358\ \text{mol};\quad n(\text{Al})=\frac{0.3358}{3}=0.1119\ \text{mol} \] \[ \text{mass}=0.1119\times27=3.02\ \text{g} \]Download The App On Google Playstore
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