Question

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

Answer 1

Answer:

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

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

Explanation:

We have to first state the fact that the reaction having the most positive reduction potential occurs at the cathode in any spontaneous electrochemical cell. The half reaction with the less positive electrode potential usually occurs at the anode.

The overall reaction equation is;

2Cu2+ (aq) + Br2 (l) ----->2Cu+ (aq) + 2Br- (aq)

E°cell= E°cathode - E°anode

E°cathode= 1.08 V

E°anode= 0.15V

E°cell = 1.08-0.15 = 0.93 V

But

∆G°= -nFE°cell

n= 2, F=96500C, E°cell= 0.93V

∆G° = -(2× 96500× 0.93)

∆G= -179490 J

But;

∆G = -RTlnK

R=8.314 JK-1

T= 25+273= 298K

Kc= the unknown

∆G° = -179490 J

Substituting values and making lnK the subject of the formula

lnK= ∆G/-RT

lnK= -( -179490/8.314 × 298)

lnK= 72.45

K= e^72.45

K= 2.91×10^31