Question

Question: An Aqueous solution is simultaneously 0.1 M in SnCl₂ and in CoCl₂

a) If the solution is electrolyzed, which metal will appear first?

b) At what decomposition potential will that metal first appear?

c) As the electrolysis proceeds, the concentration of the metal being reduced they drop, and the potential will change. How complete a separation of the metals using electrolysis it's theoretically possible? In other words at the point where the second metal begins to form, what fraction of first metal is left in the solution?

Short answer

a) “Sn” metal will appear first.

b) decomposition potential for the first metal that appears is -0.981 V.

c) fraction of first metal remains in solution is 10^-4.

Step-by-step solution

The appearance of the first metal

• Solution contains 0.10 M SnCl₂ and CoCl₂
• Standard reduction potential of Sn²⁺and Co²⁺half reactions are: -

${\text{Sn}}^{\text{2 +}}{\text{+ 2e}}^{\text{-}}\to \text{Sn(s)}$ E⁰= -0.136 V

${\text{Co}}^{2+}+2e\to Co\left(s\right)$ E⁰= -0.28 V

Reduction potential of Sn²⁺/8n is greater than reduction potential of Co²⁺/Co i.e.

${\mathrm{E}}_{{\mathrm{Sn}}^{2+}/8\mathrm{n}}^0{>\mathrm{E}}_{{\mathrm{Co}}^{2+}/\mathrm{Co}}^0$

Metal with high production potential reduced easily. So, Sn²⁺ions are reduced faster than Co²⁺ ions.

Hence if the solution is electrolyzed Sn metal will appear first.

(b): Decomposition potential for the first metal

Cathode half of the reaction: - Sn²⁺+ 2e^- = Sn (s)

${\text{Sn}}^{\text{2 +}}\left(\text{aq}\right){\text{+ 2e}}^{\text{-}}\to \text{Sn}\left(\text{s}\right)$

Reduction potential for $S{n}^{2+}/Sn$ is: -

$$\begin{gathered} E_{S{n}^{2+}/Sn}^0=E_{S{n}^{2+}/Sn}^0-\frac{0.0592}{2}log\left[\frac{Sn}{S{n}^{2+}}\right] \\ {E}_{S{n}^{2+}/Sn}^0=-0.136-0.0296log\left[\frac{1}{0.1}\right] \\ {E}_{S{n}^{2+}/Sn}^0=-0.136-0.0296 \\ {E}_{S{n}^{2+}/Sn}^0=-0.1656V \end{gathered}$$

Anode half of reaction: -

$\frac{1}{2}C{l}_2+e^{-}\rightleftharpoons C{l}^{-}$

Reduction potential for / is: -

_{role="math" localid="1663588287592" $$\begin{gathered} E_{C{l}_2/C{l}^{-}}^0=E_{C{l}_2/C{l}^{-}}^0-\frac{0.0592}{2}log\left[\frac{C{l}^{-}}{{\left(P_{C{l}_2}\right)}^{\frac{1}{2}}}\right] \\ {E}_{C{l}_2/C{l}^{-}}^0=1.36-0.0296log\left[{10}^{-1}\right] \\ {E}_{C{l}_2/C{l}^{-}}^0=1.36+0.0296 \\ {E}_{C{l}_2/C{l}^{-}}^0=1.39V \end{gathered}$$}

Hence reduction potential of anode half-cell is 1.39 V

$\begin{array}{rcl}{E}_{cell}& =& E_{cathode}^0-E_{anode}^0\\ & =& \left(-0.1656-0.815\right)V\\ & =& -0.981V\end{array}$

Thus, decomposition reaction potential for first metal appear is -0.981 V

(c): Calculate net cell potential

The electrolysis continues ${\text{Sn}}^{\text{2 +}}$iron's concentration decreases, and it ultimately decreases the potential of cathode cell, and decomposition potential increases.

If data-custom-editor="chemistry" ${\text{Sn}}^{\text{2 +}}$ concentration reaches to 10^-5, then the cathode potential changes:-

${\text{E}}_{{\text{Sn}}^{\text{2 +}}\text{/8n}}^{\text{0}}{\text{= E}}_{{\text{Sn}}^{\text{2 +}}\text{/8n}}^{\text{0}}\text{-}\frac{\text{0.0592}}{\text{2}}\text{log}\left[\frac{\text{Sn}}{{\text{Sn}}^{\text{2 +}}}\right]$

${\text{E}}_{{\text{Sn}}^{\text{2 +}}\text{/8n}}^{\text{0}}\text{= 0.136 - 0.0296log}\left[\frac{\text{1}}{{\text{10}}^{\text{- 5}}}\right]$

$$\begin{gathered} {\text{E}}_{{\text{Sn}}^{\text{2 +}}\text{/8n}}^{\text{0}}\text{= 0.136 - 0.0296×5} \\ {\text{E}}_{{\text{Sn}}^{\text{2 +}}\text{/8n}}^{\text{0}}\text{= 0.284V} \end{gathered}$$

Reduction potential of ${\text{E}}_{{\text{Co}}^{\text{2 +}}\text{/Co}}^{\text{0}}$is -0.26 V

Therefore, Co²⁺ ion starts first to reduce

• Net cell potential is: -

E_cell=

$$\begin{gathered} {\text{E}}_{\text{cathode}}^{\text{0}}{\text{- E}}_{\text{anode}}^{\text{0}} \\ \left(\text{- 0.284 - 0.815}\right)\text{V} \\ \text{- 1.09V} \end{gathered}$$

(c): remaining fraction of first metal in solution.

When decomposition potential reaches to 1.09 V then Co metals will appear. But complete separation is not possible, because cobalt start reduction when concentration of ${\text{Sn}}^{\text{2 +}}$ reaches to 10^-5

Fraction of ${\text{Sn}}^{\text{2 +}}$in solution = remaining concentration of /initial concentration of${\text{Sn}}^{\text{2 +}}$

Fraction of ${\text{Sn}}^{\text{2 +}}$in solution is

$$\begin{gathered} \text{=}\frac{{\text{10}}^{\text{- 5}}}{{\text{10}}^{\text{- 1}}} \\ {\text{= 10}}^{\text{- 4}} \end{gathered}$$

Hence, fraction of first metal remains in the solution is 10^-4.