Question

It is desired to separate and determine bismuth, copper, and silver in a solution that is 0.0550 M in BiO⁺, 0.110 M in Cu²⁺, 0.0962 M in Ag⁺, and 0.500 M in HClO₄. (a) Using 1.00 x 10^-6 M as the criterion for quantitative removal, determine whether separation of the three species is feasible by controlled-potential electrolysis. (b) If any separations are feasible, evaluate the range (versus Ag-AgCl) within which the

cathode potential should be controlled for the deposition of each.

Short answer

(a) The separation of three species is not feasible by controlled potential electrolysis.

(b) The separation is not feasible.

Step-by-step solution

Part (a) Step 1: Given Information

Whether the separation of three species is feasible by controlled potential electrolysis or not should be determined.

Part (a) Step 2: Explanation

The concentration of BiO⁺, Cu²⁺, Ag⁺, and HClO₄ is 0.0550 M, 0.110 M, 0.0962 M and 0.500 M.

The expression for electrode potential is:

${\mathrm{E}}_{\mathrm{cell}}={{\mathrm{E}}^0}_{\mathrm{cell}}-\frac{0.0592}{\mathrm{n}}\log \frac{\left[\mathrm{Red}\right]}{\left[\mathrm{Ox}\right]}-{\mathrm{E}}_{\mathrm{ref}}$…… (I)

Here, standard electrode potential is role="math" localid="1649238795640" ${E^0}_{cell}$, number of electrons is n, reduction is Red and oxidation is Ox.

The electrode potential of the cell is equal to the difference between electrode potential of cathode and electrode potential of anode.

$E_{cell}=E_{right}-E_{left}$ …… (II)

Here, the electrode potential of cathode is E_right and electrode potential of anode is E_left.

The overall reaction is:

data-custom-editor="chemistry" ${\mathrm{Ag}}^{+}+\mathrm{e}\to \mathrm{Ag}\left(\mathrm{s}\right)$

The standard electrode potential for reaction data-custom-editor="chemistry" ${\mathrm{Ag}}^{+}+\mathrm{e}\to \mathrm{Ag}\left(\mathrm{s}\right)$is 0.799 V.

The reaction at cathode is:

data-custom-editor="chemistry" ${\mathrm{Cu}}^{2+}+2\mathrm{e}\to \mathrm{Cu}\left(\mathrm{s}\right)$

The standard electrode potential for reaction data-custom-editor="chemistry" ${\mathrm{Cu}}^{2+}+2\mathrm{e}\to \mathrm{Cu}\left(\mathrm{s}\right)$is 0.337 V.

The reaction at anode is:

data-custom-editor="chemistry" ${\mathrm{BiO}}^{+}+2{\mathrm{H}}^{+}+3\mathrm{e}\to \mathrm{Bi}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}$

The standard electrode potential for reaction ${\mathrm{BiO}}^{+}+2{\mathrm{H}}^{+}+3\mathrm{e}\to \mathrm{Bi}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}$is 0.320 V.

The expression for electrode potential of cathode is:

${\mathrm{E}}_{\mathrm{cell}}={{\mathrm{E}}^0}_{\mathrm{cell}}-\frac{0.0592}{\mathrm{n}}\log \frac{1}{\left[A{g}^{+}\right]}-{\mathrm{E}}_{\mathrm{ref}}$…… (III)

The expression for electrode potential of anode is:

${\mathrm{E}}_{\mathrm{cell}}={{\mathrm{E}}^0}_{\mathrm{cell}}-\frac{0.0592}{\mathrm{n}}\log \frac{1}{\left[C{u}^{2+}\right]}-{\mathrm{E}}_{\mathrm{ref}}$…… (IV)

The expression for electrode potential bismuth (BiO⁺) is:

${\mathrm{E}}_{\mathrm{cell}}={{\mathrm{E}}^0}_{\mathrm{cell}}-\frac{0.0592}{\mathrm{n}}\log \frac{1}{\left[{\mathrm{BiO}}^{+}\right]\left[{\mathrm{H}}^{+}\right]}-{\mathrm{E}}_{\mathrm{ref}}$ …… (V)

Substitute -0.216 for E⁰_cell, 2 for n, 0.199 for E_refand 0.150 for Ag⁺ in equation (III).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=0.799V-\frac{0.0592}{2}\log \frac{1}{\left[1\times {10}^{-6}\right]}-0.199 \\ =0.444\mathrm{V}-0.199\mathrm{V} \\ =0.245\mathrm{V} \end{gathered}$$

Substitute 0.337 for E⁰_cell, 0.199 for E_ref, 1.00x10^-6 for Cu²⁺, 2 for n and in equation (IV).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=0.337V-\frac{0.0592}{2}\log \frac{1}{\left[1.00\times {10}^{-6}\right]}-0.199V \\ =0.040\mathrm{V} \end{gathered}$$

Substitute 0.320V for E⁰_cell , 0.199 for E_ref, , 1.00x10^-6 for BiO⁺, 0.5 M for H⁺ and 3 for n and in equation (V).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=0.320\mathrm{V}-\frac{0.0592}{3}\log \frac{1}{\left[1.00\times {10}^{-6}\right]{[0.5\mathrm{M}]}^2}-0.199\mathrm{V} \\ =0.009\mathrm{V} \end{gathered}$$

Therefore, the separation of three species is not feasible by controlled potential electrolysis.

Part (b) Step 1: Given Information

The range within which the cathode potential should be controlled for the deposition if separation is feasible needs to be determined.

Part (b) Step 2: Explanation

The concentration of BiO⁺, Cu²⁺, Ag⁺, and HClO₄ is 0.0550 M, 0.110 M, 0.0962 M and 0.500 M.

Substitute 0.799 V for E⁰_{cell ,}0.199 for E_ref , 0.110 for Cu2+, 2 for n and in equation (IV).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=0.799V-\frac{0.0592}{2}\log \frac{1}{\left[0.110\right]}-0.199V \\ =0.110\mathrm{V} \end{gathered}$$

Substitute 0.320V for E⁰_cell, , 0.199 for E_ref , 0.0550 for BiO⁺, 0.5 M for H⁺ and 3 for n and in equation (V).

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=0.320\mathrm{V}-\frac{0.0592}{3}\log \frac{1}{\left[0.0550\right]{[0.5\mathrm{M}]}^2}-0.199\mathrm{V} \\ =0.084\mathrm{V} \end{gathered}$$

The silver is separated from bismuth and copper, if the cell potential is greater than 0.111 V. Here, it is not greater so copper and bismuth are not separated by controlled electrode electrolysis.