Chapter 19: Q101CP (page 814)

Consider the following data:
$\begin{array}{l} {Co}^{3 +} + e^{-} \to {Co}^{2 +} E^{o} = 1.82 V \\ Co {(en)}_{3}^{2 +} K_{f} = 1.5 \times 10^{12} \\ Co {(en)}_{3}^{3 +} K_{f} = 2.0 \times 10^{47} \end{array}$
Where, en 5 ethylenediamine.
a. Calculate $E^{o}$ for the half-reaction
$Co {(en)}_{3}^{3 +} + e^{-} \to Co {(en)}_{3}^{2 +}$
b. Based on your answer to part a, which is the stronger oxidizing agent, ${Co}^{3 +}$ or $Co {(en)}_{3}^{3 +}$ ?
c. Use the crystal field model to rationalize the result in part b.

Short Answer

Expert verified

-0.26V.
Co³⁺.
It is more favorable for $Co {(H_{2} O)}_{6}^{3 +}$ than $Co {(en)}_{3}^{3 +}$ to gain an electron.

Step by step solution

Determining the standard potential for the half reaction

The given half reactions case:

$C o^{3 +} + 3 e n \to C o {(e n)}_{3}^{3 +} (K = 2.0 \times 10^{47}) C o {(e n)}_{3}^{2 +} \to C o^{3 +} + 3 e n (K = 1 / 1.5 \times 10^{12})$

The final equation is:

$C o^{3 +} + C o {(e n)}_{3}^{2 +} \to C o^{3 +} C o {(e n)}_{3}^{3 +} (\begin{matrix} K = K_{1} \times K_{2} \\ = 2.0 \times 10^{47} / 1.5 \times 10^{12} \\ = 1.3 \times 10^{35} \end{matrix}) .$

Using the Nernst equation, solve as shown below:

$E_{c e l l}^{o} = \frac{0.0591}{n} \log K .$

Put the known values in the above equation:

$\begin{matrix} E_{c e l l}^{o} = \frac{0.0591}{1} \log (1.3 \times 10^{35}) \\ = 2.08 V . \end{matrix}$

$\begin{array}{rcl} E_{CELL}^{°} & = & 1.82 - E ° . \\ 2.08 & = & 1.82 - E ° . \\ E ° & = & - 0.26 V . \end{array}$

Hence, the standard potential for the half-reaction is - 0.26 V.

Determining the stronger agent

We know that a strong oxidizing agent has more positive reduction potential and is more easily reduced.

From the reduction potentials, we can observe that Co³⁺(E° = 1.82 V) is a stronger oxidizing agent than(E° = - 0.26V).

Rationalizing the result in part (b)

In an aqueous solution, Co³⁺ forms the hydrated transition metal complex $Co {(H_{2} O)}_{6}^{3 +}$ .

In both the complexes, cobalt has a +3 oxidation state, which has 6d electrons assuming a strong field case. For each complex ion, the d orbital splitting diagram is as shown below.

d orbital splitting

When each ion complex gains an electron, it enters a higher energy eg orbital.

We know that en is a stronger field ligand than water, the d orbital crystal field splitting is larger for complex $Co {(en)}_{3}^{3 +},$ and it takes more energy to add an electron to $Co {(en)}_{3}^{3 +}$ than to $Co {(H_{2} O)}_{6}^{3 +}$ .

Hence, it is more favorable forlocalid="1663762299764" $Co {(H_{2} O)}_{6}^{3 +}$ thanlocalid="1663762310573" $Co {(en)}_{3}^{3 +}$ to gain an electron.