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
