Question

The following cell was used to determine the ${\mathrm{pSO}}_4$of a solution:

$\mathrm{SCE}\Vert {\mathrm{SO}}_4{}^{2-}(x\mathrm{M}),{\mathrm{Hg}}_2{\mathrm{SO}}_4\left({\mathrm{sat}}^2\mathrm{d}\right)\mid \mathrm{Hg}$

Calculate the ${\mathrm{pSO}}_4$ if the potential was$-0.474\mathrm{V}$

Short answer

A saturated calomel electrode (SCE) is a calomel electrode that has been saturated. As a reference electrode, it is employed. The negative cell potential is reported as $-0.474\mathrm{V}$. As a result, the reduction reaction takes place at SCE.

Step-by-step solution

Step:1 Concept Introduction

Anodic and cathodic half-cell reactions have reduction potentials of $0.615\mathrm{V}$and $0.244\mathrm{V},$respectively.

The reduction potential and half-cell reaction involved in the given cell are listed below.

At anode:
Oxidation:

$2\mathrm{Hg}\left(1\right)+{\mathrm{SO}}_4{}^{2-}(\mathrm{aq})\to {\mathrm{Hg}}_2{\mathrm{SO}}_4(\mathrm{s})+2{\mathrm{e}}^{-}$

At cathode:
Reduction:

${\mathrm{Hg}}_2{\mathrm{Cl}}_2\left(1\right)+2{\mathrm{e}}^{-}\to 2\mathrm{Hg}\left(1\right)+2{\mathrm{Cl}}^{-}\left(1\right)$

Step:2 Explanation 

The overall reaction of the given cell and the Nernst equation is given below:

${\mathrm{SO}}_4{}^{2-}(\mathrm{aq})+{\mathrm{Hg}}_2{\mathrm{Cl}}_2(1)\to 2{\mathrm{Cl}}^{-}(1)+{\mathrm{Hg}}_2{\mathrm{SO}}_4(\mathrm{s})$

${\mathrm{E}}_{\text{cell}}={\mathrm{E}}_{\text{cell}}°-\frac{0.0592}{2}\log \frac{\left[{\mathrm{Hg}}_2{\mathrm{SO}}_4\right]{\left[{\mathrm{Cl}}^{-}\right]}^2}{\left[{{\mathrm{SO}}_4}^{2-}\right]\left[{\mathrm{Hg}}_2{\mathrm{Cl}}_2\right]}$

Step:3 Overall Reaction 

The activity of the solid and liquid substance is considered to be as $1$.The concentration of localid="1645809868065" ${\mathrm{Hg}}_2{\mathrm{SO}}_4$is given as localid="1645809882883" $1\mathrm{M}$.

In the given expression, substitute the concentration of each chemical.

$$\begin{gathered} {\mathrm{E}}_{\text{cell}}={\mathrm{E}}_{\text{cell}}°-\frac{0.0592}{2}\log \frac{1}{\left[{{\mathrm{SO}}_4}^{2-}\right]} \\ {\mathrm{E}}_{\text{cell}}={\mathrm{E}}_{\text{cell}}°-\frac{0.0592}{2}\log \left(-\left[{{\mathrm{SO}}_4}^{2-}\right]\right) \\ {\mathrm{E}}_{\text{cell}}={\mathrm{E}}_{\text{cell}}°-\frac{0.0592}{2}{\mathrm{pSO}}_4 \\ -0.474\mathrm{V}=-0.371\mathrm{V}-\frac{0.0592}{2}{\mathrm{pSO}}_4 \\ -0.103=-0.0296{\mathrm{pSO}}_4 \\ {\mathrm{pSO}}_4=3.48 \end{gathered}$$