Question

Quinone undergoes a reversible reduction at a voltammetric working electrode. The reaction is

(a) Assume that the diffusion coefficients for quinone and hydroquinone are approximately the same and calculate the approximate half-wave potential (versus SCE) for the reduction of hydroquinone at an RDE from a solution buffered to a $pH$of $8.0$.

(b) Repeat the calculation in (a) for a solution buffered to a $pH$of $5.0$.

Short answer

(a) The reduction of hydroquinone at an RDE from a solution buffered to a $pH$of 8.0 is -0.118 V.

(b) The reduction of hydroquinone at an RDE from a solution buffered to a $pH$of $5.0$is 0.059 V.

Step-by-step solution

Part (a) Step 1. Given information

Quinone $\left(\mathrm{Q}\right)$undergoes a reversible reduction in hydroquinone $\left(H_{2}Q\right)$on a falling mercury electrode. The reaction is:

$\mathrm{Q}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{H}}_2\mathrm{Q}$

Electrode potential of $\mathrm{Q}/{\mathrm{H}}_2\mathrm{Q}/{\mathrm{H}}^{+}$, $E_Q^0=0.599V$

$\mathrm{pH}$of the system $=8.0$

Step 2. Equation for the half -wave potential

It is known that

Electrode potential of the reference cell (SCE), $E_{\mathrm{ref}}=0.244\mathrm{V}$

The equation is

$E_{1/2}=E_Q^0-0.0592\mathrm{V}\times \mathrm{pH}-E_{\mathrm{ref}}$

Step 3. The calculation for the approximate half-wave potential

Now put the values in the above equation

$$\begin{gathered} E_{1/2}=E_Q^0-0.0592\mathrm{V}\times \mathrm{pH}-E_{\mathrm{ref}} \\ =0.599\mathrm{V}-0.0592\mathrm{V}\times 8.0-0.244\mathrm{V} \\ {\mathrm{E}}_{1/2}=-0.118\mathrm{V} \end{gathered}$$

Part (b) Step 1. Given information

Quinone $\left(\mathrm{Q}\right)$undergoes a reversible reduction in hydroquinone $\left({\mathrm{H}}_2\mathrm{Q}\right)$on a falling mercury electrode. The reaction is:

$\mathrm{Q}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{H}}_2\mathrm{Q}$

Electrode potential of ,

$\mathrm{pH}$of the system$=5.0$

Step 2. Equation for half-wave potential

$E_{1/2}=E_Q^0-0.0592\mathrm{V}\times \mathrm{pH}-E_{\mathrm{ref}}$

Step 3. Calculation

Now put the values in the above equation

$$\begin{gathered} E_{1/2}=E_Q^0-0.0592\mathrm{V}\times \mathrm{pH}-E_{\mathrm{ref}} \\ =0.599\mathrm{V}-0.0592\mathrm{V}\times 5.0-0.244\mathrm{V} \\ {\mathrm{E}}_{1/2}=0.059\mathrm{V} \end{gathered}$$