The standard reduction potential for the reaction[itex][Co(H_2O)_6]^{3+}(aq) + e^- \to [Co(H_2O)_6]^{2+} (aq) [/itex] is about 1.8 V. The standard reduction potential for the reaction[itex][Co(NH_3)_6]^{3+}(aq) + e^- \to [Co(NH_3)_6]^{2+} (aq)[/itex] is + 0.1 V. Calculate the cell potentials to show whether the complex ions, [itex][Co(H_2O)_6]^{2+} \text{and/or} [Co(NH_3)_6]^{2+},[/itex]can be oxidized to the corresponding cobalt(III) complex by oxygen.

**My answer:**

The standard reduction potential for the reduction of oxygen to water is **+1.23 V:**

[tex]O_2(g) + 4H^+(aq) + 4e^- → 2H_2O(l) E° = +1.23 V[/tex]

To determine whether the complex ions [itex][Co(H_2O)_6]^{2+}[/itex] and [itex][Co(NH_3)_6]^{2+}[/itex] can be oxidized to the corresponding cobalt(III) complex by oxygen, we need to calculate the cell potentials for the reactions.

For the reaction involving [itex][Co(H_2O)_6]^{2+}[/itex], the cell potential is calculated as follows:

[tex]E°cell = E°cathode - E°anode

= E°(\frac{O_2}{H_2O}) - E°\left(\frac{[Co(H_2O)_6]^{3+}}{[Co(H_2O)_6]^{2+}}\right)

= (+1.23 V) - (+1.8 V)

= -0.57 V[/tex]

Since the cell potential is negative, this reaction is not spontaneous and [itex][Co(H_2O)_6]^{2+}[/itex] cannot be oxidized to [itex][Co(H_2O)_6]^{3+}[/itex] by oxygen.

For the reaction involving [itex][Co(NH_3)_6]^{2+}[/itex], the cell potential is calculated as follows:

[tex]E°cell = E°cathode - E°anode

= E°(\frac{O_2}{H_2O}) - E°\left(\frac{[Co(NH_3)_6]^{3+}}{[Co(NH_3)_6]^{2+}}\right)

= (+1.23 V) - (+0.1 V)

= +1.13 V[/tex]

Since the cell potential is positive, this reaction is spontaneous and [itex][Co(NH_3)_6]^{2+}[/itex] can be oxidized to [itex][Co(NH_3)_6]^{3+}[/itex] by oxygen.

Is this answer correct?