Question

The standard reduction potential for the reaction \({(Co{({H_2}O)_6})^{3 + }}(aq) + {e^ - } \to {(Co{({H_2}O)_6})^{2 + }}(aq)\)is about 1.8 V. The reduction potential for the reaction \({(Co{(N{H_3})_6})^{3 + }}(aq) + {e^ - } \to {(Co{(N{H_3})_6})^{2 + }}(aq)\) is +0.1 V. Calculate the cell potentials to show whether the complex ions, \({(Co{({H_2}O)_6})^{2 + }}\) and/or\({(Co{(N{H_3})_6})^{2 + }}\), can be oxidized to the corresponding cobalt (III) complex by oxygen.

Short answer

We get, \(E\left( {{{\left[ {Co{{\left( {{H_2}O} \right)}_6}} \right]}^{2 + }}} \right) = - 0.06V,E\left( {{{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]}^{2 + }}} \right) = + 1.13V\)

Step-by-step solution

Redox reaction:

A redox reaction can be defined as a chemical reaction in which electrons are transferred between two reactants participating in it. It is identified by observing the changes in the oxidation states of the reacting species.

Solving redox reaction in the task:

The standard electrode potential:

\({O_2} + 4{H^ + } + 4{e^ - } \to 2{H_2}O\) E = +1.229V

Reduction is,

\({(Co{({H_2}O)_6})^{2 + }} \to {(Co{({H_2}O)_6})^{3 + }} + {e^ - }\)

Multiply the whole with 4, since in oxidation reaction have 4 electron. So we have;

\(4{(Co{({H_2}O)_6})^{2 + }} \to 4{(Co{({H_2}O)_6})^{3 + }} + 4{e^ - }\) E = +1.8V

Cancel electron on both opposite side and substract the potential energy,

\({E_{cell}} = {E_{cathode}} - {E_{anode}}\)

\(O:{O_2} + 4{H^ + } + 4{e^ - } \to 2{H_2}O\) E = +1.229V

\(R:4{\left( {Co{{\left( {{H_2}O} \right)}_6}} \right)^{2 + }} \to 4{\left( {Co{{\left( {{H_2}O} \right)}_6}} \right)^{3 + }} + 4{e^ - }\) E = +1.8V

\(4{\left( {Co{{\left( {{H_2}O} \right)}_6}} \right)^{2 + }} + {O_2} + 4{H^ + } \to 4{\left( {Co{{\left( {{H_2}O} \right)}_6}} \right)^{3 + }} + 2{H_2}O\) E = -0.6V

Since, E is negative which means this reduction is not spontaneous.

Solving for oxidation reaction:

We have oxidation reaction, but reduction is:

\({\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)^{2 + }} \to {\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)^{3 + }} + {e^ - }\)

Multiply the whole with 4, since in oxidation reaction have 4 electron. So we have;

\(4{(Co{(N{H_3})_6})^{2 + }} \to 4{(Co{(N{H_3})_6})^{3 + }} + 4{e^ - }\) E = +0.1V

Cancel electron on both opposite side and subtract the potential energy,

\(O:{O_2} + 4{H^ + } + 4{e^ - } \to 2{H_2}O\) E = +1.229V

\(4{\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)^{2 + }} \to 4{\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)^{3 + }} + 4{e^ - }\) E = +0.1V

\(4{\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)^{2 + }} + {O_2} + 4{H^ + } \to 4{\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)^{3 + }} + 2{H_2}O\) E = +1.13V

Since, E is positive which means this reduction is spontaneous.

Hence, \(E\left( {{{\left( {Co{{\left( {{H_2}O} \right)}_6}} \right)}^{2 + }}} \right) = - 0.06V,E\left( {{{\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)}^{2 + }}} \right) = + 1.13V\)