From the redox couple below, determine the temperature at which the redox couples are in equilibrium considering that the concentrations of solutions are 1.0 M and partial pressures of the gases are at 1 atm. (USE NERNST EQUATION)
From the redox couple below, determine the temperature at which the redox couples are in equilibrium considering that the concentrations of solutions are 1.0 M and partial pressures of the gases are at 1 atm. (USE NERNST EQUATION)
2. Cathode : Cu2+/Cu
The half reaction is Cu2+ + 2e- -----> Cu, Eo = +0.34 V
Anode loses 4 electrons, so, cathode must be multiplied by '2'.
The overall reaction (with fixed anode) becomes :
2Cu2+ + 4OH- ------> O2 + 2H2O + 2Cu
Electrons transferred = 4
Calculate Eocell as shown below:
Eocell = Eocathode - Eoanode
= (0.34 V) - (0.40 V)
= -0.06 V
For present reaction, n = 4. Calculate 'K' as shown below:
Eocell−0.06 logK K =====0.0591nlogK0.05914logK−0.0141810−0.014180.9679Eocell=0.0591nlogK-0.06 =0.05914logKlogK =-0.01418K =10-0.01418=0.9679
Calculate ΔGo as shown below:
ΔGo = -nFEocell
= -(4)*(96486 J/mol-Volt )(-0.06 Volt)
= +23156.64 J/mol
= +23.16 kJ/mol
Hence, the ΔGo, K, and Eocell of redox couple is +23.16 kJ/mol, 0.9679 and -0.06 V respectively.
The ΔGo is positive and Eocell is negative, so, it is an ELECTROLYTIC CELL.
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