Determine the temperature at which the redox couples are in equilibrium, use nernst equation E = Ecell - (RT/nF) lnQ. considering that the concentrations of solutions are 1.0 M and partial pressures of the gases are at 1 atm. Redox Couple 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
Determine the temperature at which the redox couples are in equilibrium, use nernst equation E = Ecell - (RT/nF) lnQ. considering that the concentrations of solutions are 1.0 M and partial pressures of the gases are at 1 atm.
Redox Couple
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
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