In: Chemistry
The standard cell potential, E°, for Br2(aq) + Zn(s) → 2Br-(aq) + Zn2+(aq) at 298K is 1.82V. Calculate the value of ΔG° in kJ. Use 96,485 C/mol e-. Then calculate the equilibrium constant for the data.
Br2(aq) + Zn(s) → 2Br-(aq) + Zn2+(aq) E0 = 1.82v
Zn(s) -------------------> Zn^2+ (aq) + 2e^- E0 = 0.76v
Br2(aq) + 2e^- ----------------> 2Br^- (aq) E0 = 1.06v
----------------------------------------------------------------------------------------
Br2(aq) + Zn(s) → 2Br-(aq) + Zn2+(aq) E0 = 1.82v
n = 2
ΔG° = -nE0 * F
= -2*1.82*96485
= -351205J
ΔG° = -RTlnK
-351205 = -8.314*298*2.303logK
logK = -351205/-5705.8483
logK = 61.55
K = 10^61.55 = 3.63*10^61 >>>>answer