Question

1) A constant current of 0.800 A is run through the electrolytic cell in order to produce oxygen (gas) at the anode:

2 H₂O → O₂(g) + 4 e^- + 4 H⁺

The amount (in grams) of O₂ obtained after 15.2 min of electrolysis is [X] g

2) Electrolysis of a solution of Tl³⁺ produces deposition of elemental thallium, Tl, at the cathode. The time (in min.) needed for a constant current of 1.20 A to deposit 0.500 g of Tl(s) is [X] min.

3) An electrochemical cell is described by the following line notation:

Zn(s) ǀ Zn²⁺(aq) (0.0420 M) ǀǀ Tl³⁺(aq) (9.06×10^-2 M), Tl⁺(aq) (0.0400 M) ǀ Pt(s).

Given: E°(Zn²⁺/Zn) = - 0.763 V; E°(Tl³⁺/Tl⁺) = + 1.25 V.

The potential on the cathode half-cell, E₊ =  V; the potential on the anode half-cell, E_- =  V; and the overall cell potential, E_cell =  V.  

4) Consider electrolysis of molten copper bromide, CuBr₂. What is the product of this process obtained (a) at the cathode? (b) at the anode?

1. a) Cathode: Cu(s);
(b) Anode: Br₂(l)

2. (a) Cathode: Br₂(l);

(b) Anode: Cu(s)

3. (a) Cathode: Cu²⁺;

(b) Anode: Br^-

4. (a) Cathode: Br^-;

(b) Anode: Cu²⁺

5. (a) Cathode: Cu(s);

(b) Anode: O₂(g)

6. (a) Cathode: H₂(g);

(b) Anode: Cu²⁺

Answer 1

1)
O2- ----> 1/2 O2 + 2e-

for 1 mol of O2, 4 mol of electrons are required

charge passed, Q = I*t
= 0.800 A * (15.2*60)s
= 729.6 C

1 mol of electron = 96480 C

Q = 729.6 C
= 729.6 / (96480 ) mol electrons
= 7.562*10^-3 mol electrons

mol of O2 formed = (7.562*10^-3) / 4 = 1.89*10^-3 mol of O2

molar mass of O2 = 32 g/mol

mass of O2 = mol of O2 * molar mass of O2
=1.89*10^-3 * 32
= 0.060 g

Answer: 0.060 g

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