1) A constant current of 0.800 A is run through the electrolytic cell in order t

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) Zn2+(aq) (0.0420 M) Tl3+(aq) (9.06×10-2 M), Tl+(aq) (0.0400 M) 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, Ecell =  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+

Explanation / Answer

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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