Q.15E - Chemistry (OpenStax)

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.

 

Step-by-Step Solution

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 1: 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.

Step 2: 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.

Step 3: 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, \(E\left( {{{\left( {Co{{\left( {{H_2}O} \right)}_6}} \right)}^{2 + }}} \right) =  - 0.06V,E\left( {{{\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)}^{2 + }}} \right) =  + 1.13V\)