Chapter 17, Problem 17.BE

(a)

Interpretation Introduction

Interpretation:

The cathode potential at which $\text{Sb}$ deposition will happen from ${\text{SbO}}^{\text{+}}$ has to be calculated.

Concept Introduction:

When the electric current is too small, the voltage of cell is given as

$\text{E = E(cathode)}\text{-}\text{E(anode)}$

E(cathode) is electrode’s potential which is attached to negative terminal of current source.

E(anode) is electrode’s potential which is attached to positive  terminal of current source.

Concentration Polarization:  It is the change in concentration of products and reactants at electrode’s surface unlike they are same in solution.

Answer to Problem 17.BE

The cathode potential at which $\text{Sb}$ deposition will happen from ${\text{SbO}}^{\text{+}}$ is $\text{0}\text{.028 V}$

Explanation of Solution

To determine: The cathode potential at which $\text{Sb}$ deposition will happen from ${\text{SbO}}^{\text{+}}$.

$\begin{array}{l}\text{E(cathode) =}\text{0}\text{.208 -}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}\frac{\text{1}}{{\text{[SbO}}^{\text{+}}{\text{][H}}^{\text{+}}{\text{]}}^{\text{2}}}\\ \\ \text{=}\text{0}\text{.208 -}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}\frac{\text{1}}{\text{(0}\text{.010) (1}{\text{.0)}}^{\text{2}}}\\ \\ \text{=}\text{0}\text{.169}\text{V}\\ \\ \text{E(cathode versus Ag }|\text{AgCl) =}\text{E(versus S}\text{.H}\text{.E}\text{.) - E(Ag }|\text{AgCl)}\\ \\ \text{=}\text{0}\text{.169 }\text{-}\text{0}\text{.197}\text{=}\text{0}\text{.028 V}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The percentage of ${\text{Cu}}^{\text{2+}}$ ion gets converted into copper has to be calculated.

Concept Introduction:

When the electric current is too small, the voltage of cell is given as

$\text{E = E(cathode)}\text{-}\text{E(anode)}$

E(cathode) is electrode’s potential which is attached to negative terminal of current source.

E(anode) is electrode’s potential which is attached to positive  terminal of current source.

Concentration Polarization:  It is the change in concentration of products and reactants at electrode’s surface unlike they are same in solution.

Answer to Problem 17.BE

The percentage of ${\text{Cu}}^{\text{2+}}$ ion gets converted into copper is $\text{99}\text{.998\%}$.

Explanation of Solution

To determine: The percentage of ${\text{Cu}}^{\text{2+}}$ ion gets converted into copper.

The concentration of copper ion equilibrium with copper at $\text{0}\text{.169}\text{V}$ is calculated as follows

$\begin{array}{l}{\text{Cu}}^{\text{2+}}{\text{+ 2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{ Cu}}_{\text{(s)}}{\text{E}}^{\text{o}}\text{=}\text{0}\text{.339}\\ \\ \text{E(cathode) =}\text{0}\text{.339 -}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}\frac{\text{1}}{{{\text{[Cu}}^{\text{2+}}\text{]}}^{}}\\ \\ \text{0}\text{.169}\text{=}\text{0}\text{.339 -}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}\frac{\text{1}}{{{\text{[Cu}}^{\text{2+}}\text{]}}^{}}\\ \\ {{\text{[Cu}}^{\text{2+}}}^{\text{]}}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{-6}}\text{M}\end{array}$

The percentage of copper ion not reduced is determined as

$\begin{array}{l}\text{=}\frac{\text{1}\text{.8}\text{×}{\text{10}}^{\text{-6}}}{\text{0}\text{.10}}\text{×}\text{100}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{-3}}\text{\%}\\ \\ \text{\%}\text{of}\text{reduced}\text{copper}\text{ion}\text{is}\text{99}\text{.998\%}\end{array}$