Question

Write equations for the half-reactions that occur at the anode and cathode for the electrolysis of each of the following aqueous solutions.

Part A Ni(NO3)2(aq) Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.

Part B KCl(aq) Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.

Part C CuBr2(aq) Express your answers as chemical equations separated by a comma. Identify all of the phases in your answer.

Answer 1

Ni(NO3)2 = Ni+2 + NO3-

So

Cathode possible:

2 H2O (l) + 2 e - ---> H2 (g) + 2 OH- (aq) E = -0.8277 V

Ni2+ + 2 e− ⇌ Ni(s) E = −0.25

Clearly, Nickel is most likely to reduce than H2O, so choose Ni+2 to Ni(s) in cathode

for anode:

2 H2O (l) ---> O2(g) + 4 H+ (aq) + 4 e - E = -1.229 V

NO3−(aq) + 2 H+ + e− ⇌ NO2(g) + H2O +0.80

Celarly, it is much likely to oxidized H2O to form O2(g)

so

a)

half reactions:

Ni2+ + 2 e− ⇌ Ni(s)

2 H2O (l) ---> O2(g) + 4 H+ (aq) + 4 e -

b)

K+ + e− ⇌ K(s) −2.931

2 H2O (l) + 2 e - ---> H2 (g) + 2 OH- (aq) E = -0.8277 V

Clearly, H2 is going to reduce

2 H2O (l) ---> O2(g) + 4 H+ (aq) + 4 e - E = -1.229 V

2 Cl−  ⇌ Cl2(g) + 2 e−+ -1.36

The Cl- is much likely to oxidze

so

2 Cl−  ⇌ Cl2(g) + 2 e-

2 H2O (l) + 2 e - ---> H2 (g) + 2 OH- (aq)

c)

Cu2+ + 2 e− ⇌ Cu(s) +0.337

2 H2O (l) + 2 e - ---> H2 (g) + 2 OH- (aq) E = -0.8277 V

Copper is much likely to be produced

2 Br− ⇌ Br2(aq) + 2 e− -1.0873

2 H2O (l) + 2 e - ---> H2 (g) + 2 OH- (aq)

Bromine reduces

so

Cu2+ + 2 e− ⇌ Cu(s) +0.337

2 Br− ⇌ Br2(aq) + 2 e− -1.0873