Chapter 17, Problem 17.CE

(a)

Interpretation Introduction

Interpretation:

The cathode potential required to reduce ${\text{Co}}^{\text{2+}}\text{to}\text{Co}$ 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.CE

The cathode potential required to reduce ${\text{Co}}^{\text{2+}}\text{to}\text{Co}$ is $\text{-0}\text{.700 V}$

Explanation of Solution

To determine: The cathode potential required to reduce ${\text{Co}}^{\text{2+}}\text{to}\text{Co}$.

The conversion of cobalt ion into cobalt requires certain cathode potential which s calculated as follows,

$\begin{array}{l}{\text{Co}}^{\text{2+}}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Co}}_{\text{(s)}}{\text{E}}^{\text{o}}\text{=}\text{-0}\text{.282 V}\\ \\ \text{E(cathode}\text{Vs}\text{S}\text{.H}\text{.E) =}\text{-0}\text{.282 -}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}\frac{\text{1}}{{\text{[Co}}^{\text{2+}}\text{]}}\\ \\ {\text{[Co}}^{\text{2+}}\text{]}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-6}}\text{M}\text{E}\text{=}\text{-0}\text{.459 V}\\ \\ \text{E(cathode}\text{Vs}\text{S}\text{.C}\text{.E)}\text{=}\text{-0}\text{.459}\text{-}\text{0}\text{.241}\text{=}\text{-0}\text{.700 V}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The cathode potential at which the concentration of cobalt EDTA ion is $\text{1}\text{.0}\text{μM}$  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.CE

The cathode potential at which the concentration of cobalt EDTA ion is $\text{1}\text{.0}\text{μM}$ is $\text{-0}\text{.833 V}$.

Explanation of Solution

To determine: The cathode potential at which the concentration of cobalt EDTA ion is $\text{1}\text{.0}\text{μM}$ .

The conversion of cobalt EDTA ion into cobalt requires certain cathode potential which is calculated as follows,

$\begin{array}{l}{\text{Co(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_{\text{2}}{}^{\text{2-}}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Co}}_{\text{(s)}}+{\text{2C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}{\text{E}}^{\text{o}}\text{=}\text{-0}\text{.474 V}\\ \\ \text{E(cathode}\text{Vs}\text{S}\text{.H}\text{.E) =}\text{-0}\text{.474 V -}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}\frac{{[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}]}^2}{{\text{[Co(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_{\text{2}}{}^{\text{2-}}\text{]}}-0.241\\ \\ {\text{[C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{2-}}\text{]}\text{=}0.10\text{M}\\ \\ [{\text{Co(C}}_{\text{2}}{\text{O}}_{\text{4}}{\text{)}}_{\text{2}}{}^{\text{2-}}]=\text{1}\text{.0}\text{×}{\text{10}}^{\text{-6}}\text{M}\\ \\ \text{E(cathode}\text{Vs}\text{S}\text{.H}\text{.E) =}\text{-0}\text{.474 V -}\frac{\text{0}\text{.059 16}}{\text{2}}\log 100000-0.241\\ \\ \text{E}\text{=}\text{-0}\text{.833 V}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The cathode potential at which the concentration of cobalt EDTA ion is $\text{1}\text{.0}\text{μM}$ at pH seven 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.CE

The cathode potential at which the concentration of cobalt EDTA ion is $\text{1}\text{.0}\text{μM}$ at pH seven is $\text{-1}\text{.056 V}$.

Explanation of Solution

To determine: The cathode potential at which the concentration of cobalt EDTA ion at pH seven is $\text{1}\text{.0}\text{μM}$ .

The conversion of cobalt EDTA ion into cobalt requires certain cathode potential which is calculated as follows,

$\begin{array}{l}{\text{Co}}^{\text{2+}}\text{+}{\text{2e}}^{\text{-}}\stackrel{}{\rightleftharpoons }{\text{Co}}_{\text{(s)}}{\text{E}}^{\text{o}}\text{=}\text{-0}\text{.282 V}\\ \\ \text{Formation}\text{Constant}\text{for}{\text{Co(EDTA)}}^{\text{2-}}\text{is}\text{2}\text{.8}\text{×}{\text{10}}^{\text{16}}\\ \\ {\text{K}}_{\text{f}}\text{=}\frac{{\text{[Co(EDTA)}}^{\text{2-}}\text{]}}{{\text{[Co}}^{\text{2+}}{\text{]α}}_{\text{y4-}}\text{F}}\\ \\ \text{Formation}\text{Constant}\text{of}\text{EDTA}\text{is}\text{0}\text{.10M}\\ \\ {\text{α}}_{\text{y4-}}\text{=}\text{3}\text{.8}\text{×}{\text{10}}^{\text{-4}}\\ \\ {\text{[Co(EDTA)}}^{\text{2-}}\text{]}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-6}}\text{M}\\ \\ {\text{[Co}}^{\text{2+}}\text{]}\text{=}\text{9}\text{.4}\text{×}{\text{10}}^{\text{-19}}\text{M}\\ \\ \text{E =}\text{-0}\text{.282 V -}\frac{\text{0}\text{.059 16}}{\text{2}}\text{log}\frac{\text{1}}{\text{9}\text{.4}\text{×}{\text{10}}^{\text{-19}}}\text{-}\text{0}\text{.241}\\ \\ \text{E}\text{=}\text{-1}\text{.056 V}\end{array}$