Determine \({\bf{\Delta G}}\) and \({\bf{\Delta G}}^\circ \) for each of the reactions in the previous problem.

The electrical work of a cell at standard conditions with\(\Delta {G^{\rm{o}}}\), we get

\(\begin{aligned}{l}\Delta {G^o} = - nFE_{cell}^o\\F = {\rm{ Faraday costant }}(96,500{\rm{C}}/{\rm{mol}})\end{aligned}\)

For positive value of standard cell potential, the reaction is spontaneous and it is non-spontaneous for negative value of the standard cell potential.

\({\rm{Hg}}({\rm{l}}) + {{\rm{S}}^{2 - }}({\rm{aq}},0.10{\rm{M}}) + 2{\rm{A}}{{\rm{g}}^ + }({\rm{aq}},0.025{\rm{M}}) \to 2{\rm{Ag}}({\rm{s}}) + {\rm{HgS}}({\rm{s}})\)

According to the above cell reaction silver is the cathode and mercury is the anode.

So cell potential at standard condition

\(\begin{aligned}{}& = E_{{\rm{cathode }}}^0 - E_{{\rm{anode }}}^0\\ &= \{ 0.80 - ( - 0.70)\} {\rm{V}}\\ &= 1.50\;{\rm{V}}\end{aligned}\)

Cell potential at the given condition is obtained using Nernst's equation.

So we can write,

\(\begin{aligned}{}{{\rm{E}}_{{\rm{cell }}}} &= {{\rm{E}}^0}{\rm{ cell }} - (0.059/2) \times \log \left\{ {1/\left( {0.1 \times {{0.025}^2}} \right)} \right\}\\ &= 1.50\;{\rm{V}} - 0.0295 \times \log 160\;{\rm{V}}\\ &= (1.50 - 0.0295 \times 2.204){\rm{V}}\\ &= (1.50 - 0.0650){\rm{V}}\\ &= 1.435\;{\rm{V}}\end{aligned}\)

As standard cell potential and potential at given conditions are positive so the reaction will be spontaneous.\(\Delta {G^0} = - n \times F \times E_{cell}^0 = - 2 \times 96500 \times 1.5\;{\rm{J}} = - 289.5\;{\rm{kJ}}\)

\(\Delta G = n \times F \times {E_0}cell = - 2 \times 96500 \times 1.435\;{\rm{J}} = - 276.955\;{\rm{kJ}}\)

\(2{\rm{Al}}({\rm{s}}) + 3{\rm{N}}{{\rm{i}}^{ + 2}}({\rm{aq}},0.25{\rm{M}}) \to 2{\rm{A}}{{\rm{l}}^{3 + }}({\rm{aq}},0.015{\rm{M}}) + 3{\rm{Ni}}({\rm{s}})\)

According to the above cell reaction nickel is the cathode and aluminium is the anode.

So cell potential at standard condition

\(\begin{aligned}{} &= E_{{\rm{cathode }}}^0 - E_{{\rm{anode }}}^0\\ &= \{ - 0.257 - ( - 1.662)\} V\\ &= 1.405\;{\rm{V}}\end{aligned}\)

Cell potential at the given condition is obtained using Nernst's equation.

So we can write,

\(\begin{aligned}{}{E_{{\rm{cell }}}} &= E_{{\rm{cell }}}^0 - (0.059/6) \times \log \left( {{{0.015}^2}/{{0.25}^3}} \right)\\ &= \{ 1.405 - 0.0098 \times \log (225/15625)\} V\\ &= \{ 1.405 - 0.0098 \times \log 0.0144\} V\\ &= \{ 1.405 - 0.0098 \times - 1.8416\} V\\ = \{ 1.405 + 0.01804\} V\\& = 1.423\;{\rm{V}}\end{aligned}\)

As standard cell potential and potential at given conditions are positive so the reaction will be spontaneous.

\(\Delta G = n \times F \times {E_0}cell = - 2 \times 96500 \times 1.435\;{\rm{J}} = - 276.955\;{\rm{kJ}}\)\(\Delta {G^0} = - n \times F \times E_{cell}^0 = - 6 \times 96500 \times 1.405\;{\rm{J}} = - 813.495\;{\rm{kJ}}\)\(\Delta G = - n \times F \times {E_{cell}} = - 6 \times 96500 \times 1.423\;{\rm{J}} = - 823.917\;{\rm{kJ}}\)

\(6{\rm{B}}{{\rm{r}}^ - }({\rm{aq}},1.0{\rm{M}}) + 2{\rm{A}}{{\rm{l}}^{ + 3}}({\rm{aq}},0.023{\rm{M}}) \to 3{\rm{B}}{{\rm{r}}_2}({\rm{l}},0.11{\rm{M}}) + 2{\rm{Al}}({\rm{s}})\)

According to the above cell reaction aluminium is the cathode and bromine is the anode. So cell potential at standard condition

\(\begin{aligned}{} &= E_{{\rm{cathode }}}^0 - E_{{\rm{anode }}}^0\\ &= \{ - 1.6620 - 1.0873\} {\rm{V}}\\ &= 2.7493\;{\rm{V}}\end{aligned}\)

Cell potential at the given condition is obtained using Nernst's equation.

So we can write,

\(\begin{aligned}{}{{\rm{E}}_{{\rm{cell }}}} &= {{\rm{E}}^0}{\rm{ cell }} - (0.059/6) \times \log \left\{ {0.11/\left( {{{1.0}^6} \times {{0.023}^2}} \right)} \right\}\\ &= \{ - 2.7496 - 0.0098 \times \log (11000/529)\} {\rm{V}}\\ &= \{ - 2.7493 - 0.0098 \times \log 20.7939\} {\rm{V}}\\ &= \{ - 2.7493 - 0.0098 \times 1.3179\} {\rm{V}}\\ &= \{ - 2.7493 - 0.0129\} {\rm{V}}\\ &= - 2.7622\;{\rm{V}}\end{aligned}\)

As standard cell potential and potential at given conditions are negative so the

Reaction will be non-spontaneous.

\(\Delta {G^0} = - n \times F \times E_{cell}^0 = - 6 \times 96500 \times - 2.749\;{\rm{J}} = 1591.671\;{\rm{kJ}}\)\(\Delta G = - n \times F \times {E_{cell}} = - 6 \times 96500 \times - 2.7622\;{\rm{J}} = 1599.313\;{\rm{kJ}}\)