Use the data in Appendix \({\rm{L}}\) to determine the equilibrium constant for the following reactions. Assume 298.15\({\rm{K}}\) if no temperature is given.

(a) \({\bf{AgCl(s)}}\rightleftharpoons {\bf{A}}{{\bf{g}}^{\bf{ + }}}{\bf{(aq) + C}}{{\bf{l}}^{\bf{ - }}}{\bf{(aq)}}\)

(b) \({\bf{CdS(s)}}\rightleftharpoons {\bf{C}}{{\bf{d}}^{{\bf{2 + }}}}{\bf{(aq) + }}{{\bf{S}}^{{\bf{2 - }}}}{\bf{(aq)}}\) at \({\bf{377\;K}}\)

(c) \({\bf{H}}{{\bf{g}}^{{\bf{2 + }}}}{\bf{(aq) + 4B}}{{\bf{r}}^{\bf{ - }}}{\bf{(aq)}}\rightleftharpoons {\left[ {{\bf{HgB}}{{\bf{r}}_{\bf{4}}}} \right]^{{\bf{2 - }}}}{\bf{(aq)}}\)

(d) \({{\bf{H}}_{\bf{2}}}{\bf{O(l)}}\rightleftharpoons {{\bf{H}}^{\bf{ + }}}{\bf{(aq) + O}}{{\bf{H}}^{\bf{ - }}}{\bf{(aq)}}\) at \({\bf{2}}{{\bf{5}}^{\bf{^\circ }}}{\bf{C}}\)

Equilibrium constant is the ratio of products to that of reactants. It gives idea about the progress of a reversible reaction.

\({\rm{AgCl}}({\rm{s}})\rightleftharpoons {\rm{A}}{{\rm{g}}^ + } + {\rm{C}}{{\rm{l}}^ - }\)

Equilibrium constant is \(1.66 \times {10^{ - 10}}\).

Using Nernst equation we can write that equilibrium constant \({\rm{K = 1}}{{\rm{0}}^{{\rm{n}}{{\rm{E}}^{\rm{0}}}}}_{{\rm{cell}}}{\rm{/0}}{\rm{.059}}\)

\(\begin{aligned}{}{\rm{AgCl}}({\rm{s}}) + {{\rm{e}}^ - } \to {\rm{A}}{{\rm{g}}^ + } + {\rm{C}}{{\rm{l}}^ - }\left( {{{\rm{E}}^0} = 0.22233\;{\rm{V}}} \right)\\{\rm{Ag}} \to {\rm{A}}{{\rm{g}}^ + } + {{\rm{e}}^ - }\left( {{{\rm{E}}^0} = - 0.7996\;{\rm{V}}} \right)\end{aligned}\)

Adding these two equations we get \({\rm{AgCl}}({\rm{s}}) \to {\rm{A}}{{\rm{g}}^ + }({\rm{aq}}) + {\rm{C}}{{\rm{l}}^ - }({\rm{aq}})\) which has cell potential \( = (0.22233 - 0.7996){\rm{V}} = - 0.57727\;{\rm{V}}\)

Thus, equilibrium constant can be calculated as follows:

\(\begin{aligned}{}K &= 1{0^{n{E^0}_{cell}/0.059}}\\ &= 1{0^{ - 1 \times 0.57727/0.059}}\\& = 1{0^{ - 9.779}}\\ &= 1.66 \times 1{0^{ - 10}}\end{aligned}\)

\({\rm{CdS}}({\rm{s}}) \to {\rm{C}}{{\rm{d}}^{2 + }}({\rm{aq}}) + {{\rm{S}}^{2 - }}({\rm{aq}})({\rm{ At377K) }}\)

Equilibrium constant is \(2.75 \times {10^{ - 21}}\).

Using Nernst equation we can write that equilibrium constant

\({\mathop{\rm K}\nolimits} = 1{0^{n{E^0}}}_{cell }\left\{ {\{ (2.303 \times 8.314 \times 377)/96500\} = 1{0^{n{E^0}}}} \right.\) cell \(/.0746\)

\(\begin{aligned}{}CdS + 2{e^ - } \to Cd + {S^{2 - }}\left( {{E^0} = - 1.1700V} \right)\\Cd \to C{d^{2 + }} + 2{e^ - }\left( {{E^0} = 0.4030V} \right)\end{aligned}\)

Adding these two equations we get \({\mathop{\rm CdS}\nolimits} \to C{d^{2 + }} + 2{e^ - }\) which has cell potential

\( = ( - 1.1700 + 0.4030)V = - 0.7670\;V\)

Thus equilibrium constant can be calculated as follows:

\(\begin{aligned}{}K& = 1{0^{n{E^0} cell /.0746}}\\ &= 1{0^{ - 1 \times 0.57727/0.059}}\\ &= 1{0^{ - 9.779}} &= 1.66 \times 1{0^{ - 10}}\\ &= 1{0^{2 \times - 0.7670/0.0746}}\\ &= 1{0^{ - 20.56}}\\ &= 2.75 \times 1{0^{ - 21}}\end{aligned}\)

\({\rm{H}}{{\rm{g}}^{2 + }}({\rm{aq}}) + 4{\rm{B}}{{\rm{r}}^ - }({\rm{aq}}) \to {\left( {{\rm{Hg}}{{({\rm{Br}})}_4}} \right)^{2 - }}({\rm{aq}})\)

Equilibrium constant is \(5.355 \times {10^{21}}\).

Using Nernst equation we can write that equilibrium constant \({\mathop{\rm K}\nolimits} = 1{0^{n{E^0}}}_{ccl}/0.059\)

\({\mathop{\rm Hg}\nolimits} + 4B{r^ - }(aq) \to \left( {_g^H(aq) + 2{e^ - }\left( {{E^0} = - 0.21\;V} \right)} \right.\)

\({{\mathop{\rm Hg}\nolimits} ^{2 + }}(aq) + 2{e^ - } \to Hg\left( {{E^0} = 0.851\;V} \right)\)

Adding these two equations we get \({{\mathop{\rm Hg}\nolimits} ^{2 + }}(aq) + 4B{r^ - }(aq) \to {\left( {Hg{{(Br)}_4}} \right)^{2 - }}(aq)\) which has cell potential

\((0.851 - 0.210)V = 0.641\;V\)

Thus equilibrium constant can be calculated as follows:

\({\mathop{\rm K}\nolimits} = 1{0^{n{E_{cell }} / 0.059}} = 1{0^{2 \times 0.641/0.059}} = 1{0^{21.728}} = 5.355 \times 1{0^{21}}\)

\({{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to {{\rm{H}}^ + }({\rm{aq}}) + {\rm{O}}{{\rm{H}}^ - }({\rm{aq}})\)

Equilibrium constant is \(9.35 \times {10^{ - 15}}\).

Explanation of Solution

Using Nernst equation we can write that equilibrium constant \({\mathop{\rm K}\nolimits} = 1{0^{n{E^0}}}_{cell }/0.059\)

\(\begin{aligned}{}2{{\rm{H}}_2}{\rm{O}} + 2{{\rm{e}}^ - } \to {{\rm{H}}_2} + 2{\rm{O}}{{\rm{H}}^ - }\left( {{{\rm{E}}_0} = - 0.8277\;{\rm{V}}} \right)\\2{{\rm{H}}_2}{\rm{O}} + {{\rm{H}}_2} \to 2{{\rm{H}}_3}{{\rm{O}}^ + }({\rm{aq}}) + 2{\rm{O}}{{\rm{H}}^ - }({\rm{aq}})\left( {{{\rm{E}}_0} = 0.00\;{\rm{V}}} \right)\end{aligned}\)

Add these two equations.

The equation is given by:

\(2{{\rm{H}}_2}{\rm{O}} \to {{\rm{H}}_3}{{\rm{O}}^ + }({\rm{aq}}) + {\rm{O}}{{\rm{H}}^ - }({\rm{aq}})\)

Cell potential is equal to

\( - (0.8277 - 0)V = - 0.8277\;V\)

Thus equilibrium constant can be calculated as follows:

\({\mathop{\rm K}\nolimits} = 1{0^{nE_{cell }^0/0.059}} = 1{0^{1 \times - 0.8277/0.059}} = 1{0^{14.028}} = 9.35 \times 1{0^{ - 15}}\)