RI 22b. Electrochemistry 2 Tutorial (answers for self-check)

1 Suggested Solutions to Self-check for Tut 22b Electrochemistry 2 1 At the cathode(–), 2 reduction reactions are possible: 2H 2O + 2e ⇌ H2 + 2OH E = 0.83 V Ni 2+ + 2e ⇌ Ni E = 0.25 V Since E(Ni2+/Ni) is less negative than E(H2O/H2), Ni2+ is preferentially reduced to form Ni(s) at the cathode. Ni will be coated on the cathode and the cathode will increase in size / mass. At the anode(+), 3 oxidation reactions are possible. (Note: Oxidation reactions are in the reverse direction, and the starting materials that are required are on the right-hand side.) O 2 + 4H+ + 4e ⇌ 2H2O E = +1.23 V Br 2 + 2e ⇌ 2Br E = +1.07 V Ni 2+ + 2e ⇌ Ni E = 0.25 V Since E(Ni2+/Ni) is the least positive, Ni metal is preferentially oxidised to form Ni2+ at the anode. The anode dissolves. 2 (a) Volume of Al2O3 to be deposited = 500 x 1 x 10–3 = 0.5 cm3 Mass of A l2O3 = density x volume = 4.0 x 0.5 = 2.0 g Amt of A l2O3 = 2.0 / (27.0 x 2 + 16.0 x 3) = 0.0196 mol At the anode: 2H 2O(l)  O2(g) + 4H+(aq) + 4e A l reacts with O2 produced: 4Al(s) + 3O2(g)  2Al2O3(s) Amt of e – required = 4 x amt of O2 produced = 4 x ( x amt of Al2O3) = 0.1176 mol Quantity of charge needed = 96500 x 0.1176 = 11300 C Note: Al2O3 is produced by O2 formed at anode reacting with the Al anode. Al  Al3+(aq) + 3e does not occur at the anode. (b) Ni2+(aq) + 2e  Ni(s) 2Ag+(aq) + 2e  2Ag(s) For cells in series, the current passed must be the same. For the same current passed, amt of Ag: amt of Ni deposited = 2 : 1 Hence mass of Ag: mass of Ni deposited = 2 x 107.9 : 1 x 58.7 = 3.68 : 1 (c) cathode: Cu 2+(aq) + 2e  Cu(s) anode: 2H 2O(l)  O2(g) + 4H+(aq) + 4e Amt of O2 produced = ½ x amt of Cu deposited = ½ x (0.635 / 63.5) = 0.00500 mol Vol. of O 2 produced at s.t.p. = 0.00500 x 22.7 = 0.114 dm3 3 (Answer: D) Q = It = neF. Since I and F are constant, decreasing the time taken will decrease the amount of electrons that passed through the electrolyte. Applying Faraday’s First Law, since the mass of Cu deposited is proportional to the amount of electrons passed through the electrolyte, but independent of concentration of solution and nature of anion, only statement 1 is correct. 2 3

2 4 (Answer: D) The two relevant equations are: 2C l  Cl2 + 2e ---------------------------(1) 2H2O  O2 + 4H+ + 4e ------------------(2) Since V O2 = 200 cm3 and VCl2 = 100 cm3, nO2 produced = 2 x nCl2 produced. Equation (2) x 2: 4H2O  2O2 + 8H+ + 8e ------------------(3) Comparing (1) and (3), ne required to form 200 cm3 of O2 = 4 x ne to form 100 cm3 of Cl2. Since Q = It = neF, I  ne (since t and F are constant). Hence, current used in electrolysis 2 = 4 x current used in electrolysis 1 = 4 I 5 ( Answer: B) n(Al) = 0.27/27 = 0.01 mol. n(e−) = 0.03 mol. charge per mol is 2904/0.03 = 96800 C. elementary charge (value) = 1.6  10−19 C. Hence NA = 96800 ÷ 1.6  10−19 = 6.05  1023