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

a) Zn(*s*) + Pb^{2+}(*aq*) → Zn^{2+}(*aq*) + Pb(*s*)

b) Sn^{2}^{+}(*aq*) + Mg(*s*) → Sn(*s*) + Mg^{2}^{+}(*aq*)

c) Br_{2}(*l*) + 2I^{–}(*aq*) → 2Br^{–}(*aq*) + I_{2}(*s*)

d) 2Al(*s*) + 3Cu^{2}^{+}(*aq*) → 2Al^{3}^{+}(*aq*) + 3Cu(*s*)

e) O_{2}(*g*) + 4H^{+}(*aq*) + Cu(*s*) → Cu^{2+}(*aq*) + 2H_{2}O(*l*)

f) MnO_{2}(*s*) + 4H^{+}(*aq*) + Cd(*s*) → Mn^{2}^{+}(*aq*) + 2H_{2}O(*l *) + Cd^{2}^{+}(*aq*)

Δ*G*° is correlated to *E*^{o}_{cell} with the following formula:

Δ*G*° = –*nFE ^{o}*

_{cell}

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 Δ*G*°.

There are **two equations** we can use to **calculate the equilibrium constant**. One is linked through the *E*^{o}_{cell, }and the other, through the Δ*G*° which is also calculated through the *E*^{o}_{cell. }Yes, we can calculate the Δ*G*° 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), *E*^{o}_{cell }= 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 (Δ*G*°) by this formula:

Δ*G*° = –*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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