A dilute solution of sodium chloride is electrolysed using copper electrodes. (i) State one common observation at the cathode and anode. (ii) Write an equation for the reaction at each electrode. Cathode: Anode: (iii) A steady current of 2A is passed through the electrolyte for 1 hour 15 minutes. Calculate the quantity of electricity in Coulombs. (iv) Calculate the number of faradays of electricity used. (v) What volume of the gas will be liberated at the cathode at r.t.p.?

Question

Asked on March 20, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-03-20T14:57:06.461Z

(a) (i) Neutralisation reaction. (a) (ii) CuO(s) + H_2SO_4(aq) CuSO_4(aq) + H_2O(l) (a) (iii) CuO(s) + 2H^+(aq) Cu^2+(aq) + H_2O(l) (b) (i) Cathode: Reddish-brown deposit of copper forms on the cathode. Anode: The copper anode dissolves and becomes thinner (no gas is liberated). (b) (ii) Cu^2+(aq) + 2e^- Cu(s) (c) (i) Quantity of electricity: Formula: Q = I × t Given: I = 2~A, t = 1~hour = 3600~s Q = 2 × 3600 = 7200~C (c) (ii) Mass deposited on cathode: Formula: m = (M × Q)/(n × F) where M = 63.5~g~mol^-1 (molar mass of Cu), n = 2, F = 96500~C~mol^-1 First, calculate number of moles of electrons: Number of moles of e^- = (Q)/(F) = (7200)/(96500) = 0.0746~mol Moles of Cu deposited = (0.0746)/(2) = 0.0373~mol (since 2 electrons per Cu) m = 0.0373 × 63.5 = 2.37~g (c) (iii) No gas is liberated at the anode because the copper anode dissolves. Volume of gas at STP = 0~cm^3 0 cm³

Accepted Answer · 2026-03-20T14:57:06.461Z

(a) (i) Neutralisation reaction. (a) (ii) CuO(s) + H_2SO_4(aq) CuSO_4(aq) + H_2O(l) (a) (iii) CuO(s) + 2H^+(aq) Cu^2+(aq) + H_2O(l) (b) (i) Cathode: Reddish-brown deposit of copper forms on the cathode. Anode: The copper anode dissolves and becomes thinner (no gas is liberated). (b) (ii) Cu^2+(aq) + 2e^- Cu(s) (c) (i) Quantity of electricity: Formula: Q = I × t Given: I = 2~A, t = 1~hour = 3600~s Q = 2 × 3600 = 7200~C (c) (ii) Mass deposited on cathode: Formula: m = (M × Q)/(n × F) where M = 63.5~g~mol^-1 (molar mass of Cu), n = 2, F = 96500~C~mol^-1 First, calculate number of moles of electrons: Number of moles of e^- = (Q)/(F) = (7200)/(96500) = 0.0746~mol Moles of Cu deposited = (0.0746)/(2) = 0.0373~mol (since 2 electrons per Cu) m = 0.0373 × 63.5 = 2.37~g (c) (iii) No gas is liberated at the anode because the copper anode dissolves. Volume of gas at STP = 0~cm^3 0 cm³

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