PROBLEMA Nickel-iron battery is composed by a Fe anode and a NiO cathode, in solution with the electrolyte KOH. The battery has a standard cell potential of Δε
cell°= 0.400 V and operates through the following half reactions, at T= 25°C:
Fe + 2OH
- Fe(OH)
2 + 2e
-NiO + H
2O + e
- Ni(OH)
2 + OH
-Determine,
- ΔG° of the reaction and Keq.
- E.M.F. when [KOH]= 5.00 M
ATTEMPTFor the first point I know that ΔG°= -nΔε
cell°F. From the half reactions I notice that n=2
=> ΔG°= -(2)(0.400 J/C)(96485 C/mol)= -7.72 10
4 J/mol.
From this I can then determine K
eq.=[itex]e^{(\frac {-\Delta G°}{RT})}[/itex] = [itex]e^{(\frac {(7.72)(10^{4} J/mol)}{(8.31 J/(mol K))(298 K)})}[/itex]= 31.2.
For the second point E.M.F. should ≡ Δε
cell = [itex]\Deltaε_{cell}° + \frac {RT}{nF}\ln Q[/itex]. What I don't understand is how to determine Q. I know that Q is the reaction quotient so I thought of analyzing the total reaction of the battery:
Fe + 2OH
- + 2NiO + 2H
2O + 2e
- Fe(OH)
2 + 2e
- + 2Ni(OH)
2 + 2OH
-But now if I simplify, I obtain: Fe + 2NiO + 2H
2O
Fe(OH)
2 + 2Ni(OH)
2 and consequently log(Q)= 0 since Q= 1 because under the conditions of the problem all reactants and products have an activity of 1.