Question

Consider the following half reactions and their standard reduction potential values to answer the following questions.

Answer 1

Answer:

Cu2+ (aq) + e- -------->Cu+ (aq)​E° = 0.15 V anode

Br2 (l) + 2e- -------------> 2Br- (aq) ​E° = 1.08 V cathode

Explanation:

The half equation having a more positive reduction potential indicates the reduction half equation while the half equation having the less positive reduction potential indicates the oxidation half equation.

The overall redox reaction equation is;

2Cu^+(aq) + Br2(g) ----> 2Cu^2+(aq) + 2Br^-(aq)

E°cell= E°cathode -E°anode

E°cathode= 1.08 V

E°anode= 0.15 V

E°cell= 1.08V-0.15V

E°cell= 0.93V

From;

∆G=-nFE°cell

n= 2

F=96500C

E°cell= 0.93V

∆G= -(2×96500×0.93)

∆G= - 179.49 J

From ∆G= -RTlnK

∆G= - 179.49 J

R=8.314Jmol-K-1

T= 25° +273= 298K

K= ???

lnK= ∆G/-RT

lnK=-( - 179.49/(8.314×298))

lnK= 0.0724

K= e^0.0724

Kc= 1.075