Transition Elements: Question 6

Worked solution

Part (a): Instability of $[\text{Co(H}_2\text{O)}_6]^{3+}(aq)$ in water

Comparing the two given electrode potentials: $E^{\ominus}([\text{Co(H}_2\text{O)}_6]^{3+}/[\text{Co(H}_2\text{O)}_6]^{2+})=+1.82\ \text{V} > E^{\ominus}(\text{O}_2/\text{H}_2\text{O})=+1.23\ \text{V}$

The half-equation with the more positive $E^{\ominus}$ has the greater tendency to proceed as a reduction, so $[\text{Co(H}_2\text{O)}_6]^{3+}(aq)$ is reduced to $[\text{Co(H}_2\text{O)}_6]^{2+}(aq)$, while water is oxidised to $\text{O}_2$ (i.e. the second half-equation runs in reverse): $E^{\ominus}_{cell}=E^{\ominus}_{reduction}-E^{\ominus}_{oxidation}=1.82-1.23=+0.59\ \text{V}$

Since $E^{\ominus}_{cell}$ is positive, this reaction is thermodynamically feasible: $[\text{Co(H}_2\text{O)}_6]^{3+}(aq)$ is a strong enough oxidising agent to oxidise water itself, so it cannot exist for long in aqueous solution, it decomposes, being reduced to $[\text{Co(H}_2\text{O)}_6]^{2+}(aq)$ as it liberates oxygen gas.

To combine the half-equations, the one-electron cobalt half-equation is multiplied by 4 to match the four electrons in the oxygen half-equation: $4[\text{Co(H}_2\text{O)}_6]^{3+}(aq) + 4e^- \rightarrow 4[\text{Co(H}_2\text{O)}_6]^{2+}(aq)$ $2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-$

Adding and cancelling the $4e^-$: $4[\text{Co(H}_2\text{O)}_6]^{3+}(aq) + 2\text{H}_2\text{O}(l) \rightarrow 4[\text{Co(H}_2\text{O)}_6]^{2+}(aq) + \text{O}_2(g) + 4\text{H}^+(aq)$

(Charge check: LHS $4(+3)=+12$; RHS $4(+2)+0+4(+1)=+12$, balanced. O check: LHS $4\times6+2=26$; RHS $4\times6+2=26$, balanced. H check: LHS $4\times12+2\times2=52$; RHS $4\times12+4=52$. Balanced.)

Part (b): Oxidation of $[\text{Co(NH}_3\text{)}_6]^{2+}(aq)$ by atmospheric oxygen

Here the relevant comparison is reversed: $\text{O}_2/\text{H}_2\text{O}$ (in its alkaline form) now has the more positive potential, so it is $\text{O}_2$ that is reduced, and the cobalt(II) ammine complex that is oxidised: $E^{\ominus}(\text{O}_2/\text{OH}^-)=+0.40\ \text{V} > E^{\ominus}([\text{Co(NH}_3\text{)}_6]^{3+}/[\text{Co(NH}_3\text{)}_6]^{2+})=+0.10\ \text{V}$ $E^{\ominus}_{cell}=E^{\ominus}_{reduction}-E^{\ominus}_{oxidation}=0.40-0.10=+0.30\ \text{V}$

Since $E^{\ominus}_{cell}$ is positive, this oxidation is feasible: atmospheric oxygen can indeed oxidise $[\text{Co(NH}_3\text{)}_6]^{2+}(aq)$ to $[\text{Co(NH}_3\text{)}_6]^{3+}(aq)$ in ammoniacal solution.

Multiplying the reversed cobalt half-equation by 4 to match the four electrons of the oxygen half-equation: $\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq)$ $4[\text{Co(NH}_3\text{)}_6]^{2+}(aq) \rightarrow 4[\text{Co(NH}_3\text{)}_6]^{3+}(aq) + 4e^-$

Adding and cancelling the $4e^-$: $\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4[\text{Co(NH}_3\text{)}_6]^{2+}(aq) \rightarrow 4\text{OH}^-(aq) + 4[\text{Co(NH}_3\text{)}_6]^{3+}(aq)$

(Charge check: LHS $0+0+4(+2)=+8$; RHS $4(-1)+4(+3)=+8$, balanced. O check: LHS $2+2=4$; RHS $4$, balanced. H check: LHS $2\times2=4$; RHS $4\times1=4$, balanced. N check: LHS $4\times6=24$; RHS $4\times6=24$, balanced.)

Part (c): Why $\text{NH}_3$ stabilises the $+3$ oxidation state relative to $+2$

$\text{Co}^{3+}$ is smaller and carries a greater positive charge than $\text{Co}^{2+}$, so it is a much more strongly polarising cation and forms much stronger dative (coordinate) bonds with any given ligand. The key point is that this strengthening effect is not the same size for every ligand. The increase in bond strength on going from $\text{Co}^{2+}$ to $\text{Co}^{3+}$ is far greater when the ligand is $\text{NH}_3$ than when it is $\text{H}_2\text{O}$.

Consequently, replacing the six $\text{H}_2\text{O}$ ligands with six $\text{NH}_3$ ligands releases disproportionately more extra complexation energy for the $\text{Co}^{3+}$ complex than for the $\text{Co}^{2+}$ complex. This extra stabilisation of the $+3$ state relative to the $+2$ state means less energy is required overall to remove the additional electron from the ammine complex than from the aqua complex, so the standard electrode potential for the $\text{Co}^{3+}/\text{Co}^{2+}$ couple is much lower ($+0.10\ \text{V}$) with $\text{NH}_3$ ligands than with $\text{H}_2\text{O}$ ligands ($+1.82\ \text{V}$). The $\text{NH}_3$ ligand makes the $+3$ oxidation state considerably easier to access and more stable relative to $+2$.

Part (d): Coordination number and shape of $[\text{Co(NH}_3\text{)}_6]^{3+}$

Cobalt is bonded to six $\text{NH}_3$ ligands, each donating one lone pair to form one coordinate bond, so:

• Coordination number $=6$
• Shape $=$ octahedral

Final answers

• (a) $E^{\ominus}_{cell}=+0.59\ \text{V}$ (feasible); $4[\text{Co(H}_2\text{O)}_6]^{3+}(aq)+2\text{H}_2\text{O}(l)\rightarrow4[\text{Co(H}_2\text{O)}_6]^{2+}(aq)+\text{O}_2(g)+4\text{H}^+(aq)$, the aqua $\text{Co}^{3+}$ complex decomposes, oxidising water.
• (b) $E^{\ominus}_{cell}=+0.30\ \text{V}$ (feasible); $\text{O}_2(g)+2\text{H}_2\text{O}(l)+4[\text{Co(NH}_3\text{)}_6]^{2+}(aq)\rightarrow4\text{OH}^-(aq)+4[\text{Co(NH}_3\text{)}_6]^{3+}(aq)$. The ammine $\text{Co}^{2+}$ complex is readily air-oxidised.
• (c) $\text{NH}_3$ forms disproportionately stronger bonds to the smaller, more highly charged $\text{Co}^{3+}$ than to $\text{Co}^{2+}$, releasing extra complexation energy that stabilises the $+3$ state and lowers $E^{\ominus}$.
• (d) Coordination number $=6$; shape $=$ octahedral.