Question

What cathode potential (versus SCE) would be required to lower the total Hg(II) concentration of the following solutions to 1.00 x 10^-6 M (assume reaction product in each case is elemental Hg):

(a) an aqueous solution of Hg²⁺?

(b) a solution with an equilibrium SCN^- concentration of 0.100 M?

${\mathrm{Hg}}^{2+}+2{\mathrm{SCN}}^{-}\rightleftharpoons \mathrm{Hg}{\left(\mathrm{SCN}\right)}_2\left(\mathrm{aq}\right){\mathrm{K}}_{\mathrm{f}}=1.8\times {10}^7$

(c) a solution with an equilibrium Br2 concentration of 0.100 M?

${{\mathrm{HgBr}}_4}^{2-}+2\mathrm{e}\rightleftharpoons \mathrm{Hg}\left(\mathrm{l}\right)+4{\mathrm{Br}}^{-}{\mathrm{E}}^0=0.223\mathrm{V}$

Short answer

(a) The cathode potential required to lower the Hg(II) concentration of an aqueous solution of Hg²⁺ to 1.00 x 10^-6 M is 0.4324 V.

(b) The cathode potential required to lower the Hg(II) concentration of a solution with an equilibrium SCN^- to 1.00 x 10^-6 M is 0.943 V.

(c) The cathode potential required to lower the Hg(II) concentration of a solution with an equilibrium Br^- to 1.00 x 10^-6 M is -0.08 V.

Step-by-step solution

Part (a) Step 1: Given Information

The cathode potential required to lower the Hg(II) concentration of an aqueous solution of Hg²⁺ to 1.00 x 10^-6 M is to be stated.

Part (b) Step 2: Explanation

The given reaction is:

${\mathrm{Hg}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Hg}\left(\mathrm{l}\right)$…… (1)

Refer to Appendix 3 to obtain the value of electrode potential of reduction process for Equation (I) as 0.854 V.

The reaction for the oxidation process is:

${\mathrm{HgCl}}_2\left(\mathrm{s}\right)+2{\mathrm{e}}^{-}\rightleftharpoons 2\mathrm{Hg}\left(\mathrm{l}\right)+2{\mathrm{Cl}}^{-}$…… (II)

The value of electrode potential of oxidation process for Equation (II) as 0.244 V.

The expression for the cell potential is:

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{c}}-{\mathrm{E}}_{\mathrm{a}}$ …… (III)

Here, the cell potential is $E_{cell}$, the potential at cathode is $E_c$, and the potential at anode is $E_a$.

The expression the Nernst equation for half cell potential at room temperature is:

$E=E^0-\frac{0.05916V}{n}{\log }_{10}\frac{a_{red}}{a_{ox}}$ ……. (IV)

Here, the half cell potential is E, the standard half cell reduction potential is E⁰, chemical activity of reductant is a_red, chemical activity of oxidant is a_ox, and the number of moles of electrons is n.

Substitute E_c for E, 0.854 V for E⁰, 2for n, and 10^-6 for $\frac{{\mathrm{a}}_{\mathrm{ox}}}{{\mathrm{a}}_{\mathrm{red}}}$in Equation (IV).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{c}}=0.854\mathrm{V}-\frac{0.05916\mathrm{V}}{2}{\log }_{10}\frac{1}{{10}^{-6}} \\ =0.854\mathrm{V}-(-0.17748\mathrm{V}) \\ =0.6764\mathrm{V} \end{gathered}$$

Substitute 0.6764 V for E_c and 0.244 V E_a in Equation (III).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=0.6764\mathrm{V}-0.244\mathrm{V} \\ =0.4324\mathrm{V} \end{gathered}$$

Therefore, the cathode potential required to lower the Hg(II) concentration of an aqueous solution of Hg²⁺ to 1.00 x 10^-6 M is 0.4324 V.

Part (b) Step 1: Given Information

The cathode potential required to lower the Hg(II) concentration of a solution with an equilibrium SCN^- to 1.00 x 10^-6 M is to be stated.

Part (b) Step 2: Explanation

The reaction for the reduction process is given below.

${\mathrm{Hg}}^{2+}+2{\mathrm{SCN}}^{-}\rightleftharpoons \mathrm{Hg}{\left(\mathrm{SCN}\right)}_2\left(\mathrm{aq}\right)$…… (V)

The value of electrode potential of reduction process for Equation (V) as 0.854 V.

The expression for the formation constant is:

${\mathrm{K}}_{\mathrm{f}}=\frac{\left[\mathrm{Hg}{\left(\mathrm{SCN}\right)}_2\right]}{\left[{\mathrm{Hg}}^{2+}\right]{\left[{\mathrm{SCN}}^{-}\right]}^2}$…… (VI)

Here, the formation constant is K_f, the concentration of reactants is [Hg(SCN)₂], the concentration of Hg²⁺ is [Hg²⁺], and the concentration of SCN^- is [SCN^- ].

The expression for the cell potential is:

$E_{cell}=E_c-E_a$ …… (VII)

Here, the cell potential is E_cell, the potential at cathode is E_c, and the potential at anode is E_a.

The expression the Nernst equation for half cell potential at room temperature is:

$E_c=E^0-\frac{0.05916V}{n}{\log }_{10}\frac{1}{\left[H{g}^{2+}\right]}$……. (VIII)

Here, the half cell cathode potential is E_c, the standard half cell reduction potential is E⁰, the concentration of Hg²⁺ is [Hg²⁺], and the number of moles of electrons is n.

Substitute 0.100 M for [SCN^-], 1.00 x 10^-6 M for[Hg(SCN)₂] and 1.8x10⁷ for K_f in Equation (VI).

$$\begin{gathered} 1.8\times {10}^7=\frac{1.00\times {10}^{-6}\mathrm{M}}{\left[{\mathrm{Hg}}^{2+}\right]{(0.100\mathrm{M})}^2} \\ \left[{\mathrm{Hg}}^{2+}\right]=\frac{1.00\times {10}^{-6}\mathrm{M}}{{(0.100\mathrm{M})}^2\left(1.8\times {10}^7\right)} \\ \left[{\mathrm{Hg}}^{2+}\right]=5.56\times {10}^{-12}\mathrm{M} \end{gathered}$$

Substitute 1.187 V for E_c and 0.244 V E_a in Equation (VII).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=1.187\mathrm{V}-0.244\mathrm{V} \\ =0.943\mathrm{V} \end{gathered}$$

Therefore, the cathode potential required to lower the Hg(II) concentration of a solution with an equilibrium SCN^- to 1.00 x 10^-6 M is 0.943 V.

Substitute 0.854 V for E⁰, 2 for n, and 5.56x10^-12for $\frac{{\mathrm{a}}_{\mathrm{ox}}}{{\mathrm{a}}_{\mathrm{red}}}$in Equation (VIII).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{c}}=0.854\mathrm{V}-\frac{0.05916\mathrm{V}}{2}{\log }_{10}\frac{1}{5.56\times {10}^{-12}} \\ =0.854\mathrm{V}-(-0.333\mathrm{V}) \\ =1.187\mathrm{V} \end{gathered}$$

Part (c) Step 1: Given Information

The cathode potential required to lower the Hg(II) concentration of a solution with an equilibrium Br^- concentration of 0.100 M to 1.00 x 10^-6M is to be stated.

Part (c) Step 2: Explanation

The reaction for the process is:

${{\mathrm{HgBr}}_4}^{2-}+2\mathrm{e}\rightleftharpoons \mathrm{Hg}\left(\mathrm{l}\right)+4{\mathrm{Br}}^{-}$…… (IX)

The value of electrode potential of process in Equation (IX) is 0.223 V.

The expression for the cell potential is:

$E_{cell}=E_c-E_a$…… (X)

Here, the cell potential is E_cell, the potential at cathode is E_c, and the potential at anode is E_a.

The expression the Nernst equation for half cell potential at room temperature is:

$E_c=E^0-\frac{0.05916V}{n}{\log }_{10}\frac{{\left[B{r}^{-}\right]}^4}{\left[HgB{r_4}^{2-}\right]}$……. (XI)

Here, the standard half cell reduction potential is E⁰, the concentration of bromine is [Br^-], the concentration of HgBr₄^2- is [HgBr₄^2-] , and the number of moles of electrons is n.

Substitute 0.223 V for E⁰, 2 for n, 0.100 M for [Br-] and 1.00x10^-6 M for [HgBr42-] in Equation (XI).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{c}}=0.223\mathrm{V}-\frac{0.05916\mathrm{V}}{2}{\log }_{10}\frac{{\left[0.100\right]}^4\mathrm{M}}{\left[1.00\times {10}^{-6}\right]\mathrm{M}} \\ =0.854\mathrm{V}-0.05916\mathrm{V} \\ =0.16384\mathrm{V} \end{gathered}$$

Substitute 0.16384 V for E_c and 0.244 V for E_a in Equation (III).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=0.16384\mathrm{V}-0.244\mathrm{V} \\ =-0.08\mathrm{V} \end{gathered}$$

Therefore, the cathode potential required to lower the Hg(II) concentration of a solution with an equilibrium Br^- to 1.00x10^-6 is -0.08 V.