Question

(a) Calculate the standard potential for the reaction

For ${\mathrm{Ag}}_3{\mathrm{AsO}}_4,K_{sp}=1.2\times {10}^{-22}$

(b) Give a schematic representation of a cell with a silver indicator electrode and an SCE as reference that could be used for determining $\mathrm{AsO}{4}^{3-}$.
(c) Derive an equation that relates the measured potential of the cell in (b) to (assume that the junction potential is zero).
(d) Calculate the pAsO4 of a solution that is saturated with and contained in the cell described in (b) if the resulting potential is 0.193 .

Short answer

In potentiometric titrations, an indicator electrode is a type of electrode. It's a type of endpoint indication. Indicator electrodes include membrane electrodes and metallic electrodes.

(a) The reduction potential for above reaction is 0.799 V

(b) $\mathrm{SCE}\Vert {\mathrm{Ag}}_3{\mathrm{AsO}}_4\left(\text{sat'd}\right),{\mathrm{AsO}}_4{}^{3-}(x\mathrm{M})\mid \mathrm{Ag}\text{.}$

(c) As a result, the equation that relates the cell's measured potential in (b)

(d) Therefore, the value of pAsO4 is 3.60

Step-by-step solution

Step 1. Explanation of (a)

The standard potential for the given reaction

Explanation of Solution
The reaction is given as

${\mathrm{Ag}}_3{\mathrm{AsO}}_4\left(s\right)+3{\mathrm{e}}^{-}\rightleftharpoons 3\mathrm{Ag}\left(s\right)+{{\mathrm{AsO}}_4}^{3-}$

The decomposition of ${\mathrm{Ag}}_3{\mathrm{AsO}}_4$is shown below.

${\mathrm{Ag}}_3{\mathrm{AsO}}_4\left(s\right)+3{\mathrm{e}}^{-}\rightleftharpoons 3{\mathrm{Ag}}^{+}+{{\mathrm{AsO}}_4}^{3-}$

For the aforementioned reaction, the solubility product is given by the formula,

$$\begin{gathered} K_{\text{sp}}={\left[{\mathrm{Ag}}^{+}\right]}^3\left[{{\mathrm{AsO}}_4}^{3-}\right] \\ {\left[{\mathrm{Ag}}^{+}\right]}^3=\frac{K_{\text{sp}}}{\left[{{\mathrm{AsO}}_4}^{3-}\right]} \end{gathered}$$

The half-cell reduction reaction of Ag to Ag+ is given as,

$3{\mathrm{Ag}}^{+}+3{\mathrm{e}}^{-}\rightleftharpoons 3\mathrm{Ag}\left(s\right)$

The reduction potential for above reaction is 0.799 V

Step 2. Activity

Substitute the equation (1), activity of silver solid, and in the above expression for the activity of solid.

It is considered that when the concentration of silver ion 1 , then Substitute the $E_{\text{cell}}=E_{{\mathrm{Ag}}_3{\mathrm{AsO}}_4}.$in above expression.

$$\begin{gathered} E_{\mathrm{cell}}=E_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{\mathrm{o}}-\frac{0.0592}{3}\log \frac{1}{{\left[{\mathrm{Ag}}^{+}\right]}^3} \\ =0.799-\frac{0.0592}{3}\log \frac{\left[{{\mathrm{AsO}}_4}^{3-}\right]}{K_{\mathrm{sp}}} \end{gathered}$$

It is considered that when the concentration of silver ion 1 , then Substitute the in above expression.

$$\begin{gathered} E_{{\mathrm{Ag}}_3{\mathrm{AsO}}_4}=0.799-\frac{0.0592}{3}\log \frac{1}{1.2\times {10}^{-22}} \\ =0.799-0.0197\log {\left[1.2\times {10}^{-22}\right]}^{-} \\ =0.799-0.0197\left[-\log \left[1.2\times {10}^{-22}\right]\right] \\ =0.799-0.0197[-[\log 1.2-22\log 10\left]\right] \\ =0.799-0.0197[-0.079+22] \\ =0.799-0.4318 \\ =0.3664\mathrm{V} \end{gathered}$$

Step 3. Explanation of part (b)

Therefore, the standard potential for the given reaction is $0.364$V

The schematic representation of a cell with a silver indicator electrode and a reference SCE is

$\mathrm{SCE}\Vert {\mathrm{Ag}}_3{\mathrm{AsO}}_4\left(\text{sat'd}\right),{\mathrm{AsO}}_4{}^{3-}(x\mathrm{M})\mid \mathrm{Ag}\text{.}$

Oxidation and reduction are the two reactions that take place in the cell. The parallel line is employed in cell notation to differentiate these two half-cell reactions.

The SCE is a reference electrode for determining the concentration .As a result, it should be written on the cell representation left side.

Step 4. Cell representation

The reaction is given as

${\mathrm{Ag}}_3{\mathrm{AsO}}_4\left(s\right)+3{\mathrm{e}}^{-}\rightleftharpoons 3{\mathrm{Ag}}^{+}+{\mathrm{AsO}}_4^{3-}$

It's a reduction reaction, and it's written on the right side of the cell depiction.

As a result, the schematic of a cell with a silver indicator electrode and a reference SCE is shown.

$\mathrm{SCE}\Vert {\mathrm{Ag}}_3{\mathrm{AsO}}_4\left(\text{sat'd}\right),{\mathrm{AsO}}_4{}^{3-}(x\mathrm{M})\mid \mathrm{Ag}\text{.}$

Step 5. Explanation of (c)

The equation that relates the measured potential of the cell in (b) to

${\mathrm{pAsO}}_4$is

The cell is $\mathrm{SCE}\parallel {\mathrm{Ag}}_3{\mathrm{AsO}}_4$(sat'd), ${{\mathrm{AsO}}_4}^{3-}\left(x\mathrm{M}\right)\mid \mathrm{Ag}$.

The corresponding half - cell reactions are shown below :

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

Cathode : ${\mathrm{AsO}}_4^{3-}+3{\mathrm{e}}^{-}\rightleftharpoons {\mathrm{AsO}}_4$

Therefore, the reduction potential of cathodic and anodic half-cell reaction are 0.355V and 0.244 V respect.

$E_{\text{cell}}$for the reaction is given by an expression :

$E^{\circ }\text{cell}=E^{\circ }\text{cathode}-E^{\circ }\text{anode}$

Substitute reduction potential for half-cell reaction in above formula is given as :

$\begin{array}{rl}{E}_{\text{cell}}^{\circ }& =0.366-0.244\\ & =0.122\mathrm{V}\end{array}$

Therefore, the ${\mathrm{E}}^0$ cell of the cell is $0.122\mathrm{v}$

The anodic and cathodic reactions are multiplied by 3 and 2 accordingly to balance the quantity of electrons.

As a result, the general reaction is

The Nernst equation for the above reaction is represented as shown below:

$E_{\text{cell}}=E_{\text{cell}}°-\frac{0.0592}{6}\log \frac{{\left[{\mathrm{Hg}}_2{\mathrm{Cl}}_2\right]}^6{\left[{\mathrm{AsO}}_4\right]}^2}{{\left[{\mathrm{Cl}}^{-}\right]}^6[\mathrm{Hg}{]}^6{\left[{{\mathrm{AsO}}_4}^{3-}\right]}^2}$

The activity (1) of each solid and liquid substance is considered to be the Substitute concentration of each substance.

Step:6 Substitution

$$\begin{gathered} E_{\text{cell}}=0.122-\frac{0.0592}{6}\log \frac{1}{{\left[{{\mathrm{AsO}}_4}^{3-}\right]}^2} \\ {E}_{\text{cell}}=0.122-\frac{0.0592}{6}\log {\left[{{\mathrm{AsO}}_4}^{3-}\right]}^{-2} \\ =0.122-\frac{0.0592}{6}\left[-2\log \left[{{\mathrm{AsO}}_4}^{3-}\right]\right] \\ =0.122+\frac{0.0592{\mathrm{pAsO}}_4}{3} \\ {E}_{\text{cell}}-0.122=\frac{0.0592{\mathrm{pAsO}}_4}{3} \\ {\mathrm{pAsO}}_4=\frac{\left(E_{\text{cel}}-0.122\right)\times 3}{0.0592} \end{gathered}$$

As a result, the equation that relates the cell's measured potential in (b)

Step:7 Explanation of (d)

The ${\mathrm{pAsO}}_4$of a solution that is saturated with ${\mathrm{Ag}}_3{\mathrm{AsO}}_4$ and contained in the cell described in (b) if the resulting potential is $0.193\mathrm{V}$is$2.28$
Explanation of Solution
The expression for calculating the value of is derived in the part (c) as shown below.

Step:8 Substitution

The expression for calculating the value of ${\mathrm{pAsO}}_4$ is derived in the part (c) as shown below.

${\mathrm{pAsO}}_4=\frac{\left(E_{\text{cell}}-0.122\right)\times 3}{0.0592}$

The resulting potential is given as $0.193\mathrm{V}$. Substitute $E_{\text{cell}}$in above expression

$\begin{array}{r}{\mathrm{pAsO}}_4=\frac{(0.193-0.122)\times 3}{0.0592}\\ =\frac{0.213}{0.0592}\\ =3.60\end{array}$

Therefore, the value of ${\mathrm{pAsO}}_4$is$3.60$