Calculate the cell potential, Gibbs free energy change ΔG°, and equilibrium constant, K for each redox reaction:

a) Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)

b) Sn2+(aq) + Mg(s) → Sn(s) + Mg2+(aq)

c) Br2(l) + 2I(aq) → 2Br(aq) + I2(s)

d) 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

e) O2(g) + 4H+(aq) + Cu(s) → Cu2+(aq) + 2H2O(l)

f) MnO2(s) + 4H+(aq) + Cd(s) → Mn2+(aq) + 2H2O() + Cd2+(aq)


ΔG° is correlated to Eocell with the following formula:

ΔG° = –nFEocell


Where n is the moles of electrons and F is the Faraday’s constant which is equal to 96,485 C. Therefore, the plan is to first calculate the cell potential and then use it to determine the ΔG°.


There are two equations we can use to calculate the equilibrium constant. One is linked through the Eocell, and the other, through the ΔG° which is also calculated through the Eocell. Yes, we can calculate the ΔG° using the standard free energies of formations, however, this is not related to electrochemistry and we won’t focus on that here.


\[{E^o}_{{\rm{cell}}}\, = \,\frac{{0.0257\;{\rm{V}}}}{n}\,\ln \,K\]

\[K\; = \;{e^{\frac{{nE}}{{0.0257}}}}\]


For reaction (a), Eocell = 0.63 V, therefore, K would be:


\[K\; = \;{e^{\frac{{2\, \cdot \,0.63}}{{0.0257}}}}\, = \,{e^{49.0}}\, = \;1.91\, \times \,{10^{21}}\]


The second approach:

The equilibrium constant is correlated with the standard free energy change (ΔG°) by this formula:

ΔG° = –RT ln K 

Therefore, the equilibrium constant would be equal to:


\[\ln \,K\; = \;\frac{{ – \Delta G^\circ }}{{RT}}\]

\[K\; = \;{e^{\frac{{ – \Delta G^\circ }}{{RT}}}}\]

For the next problems, we will use the first approach to calculate the K.


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