Question

Calculate the cathode potential (versus S.C.F.) needed to reduce cobalt(II) to $\mathbf{1}\mathbf{.}\mathbf{0}\mathit{\mu }\mathit{M}$in each of the following solutions. In each case, Co(s) is the product of the reaction. (Disregard any overpotential.)

(a) $\mathbf{0}\mathbf{.}\mathbf{10}\mathit{M}\mathit{H}\mathit{C}\mathit{l}{\mathit{O}}_{\mathbf{4}}$

(b)$\mathbf{0}\mathbf{.}\mathbf{10}\mathit{M}{\mathit{C}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}$(Find the potential at which)$\left[Co(C_2{O}_4{)}_2^{2-}\right]\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathit{\mu }\mathit{M}$

role="math" localid="1668354628300" $\mathit{C}\mathit{o}\mathbf{(}{\mathit{C}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{2}}^{\mathbf{2}\mathbf{-}}\mathbf{+}\mathbf{2}{\mathit{c}}^{\mathbf{-}}\mathbf{\to }\mathit{C}\mathit{o}\mathbf{(}\mathit{s}\mathbf{)}\mathbf{+}\mathbf{2}{\mathit{C}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}\mathbf{\hspace{0.33em}}{\mathit{E}}^{\mathbf{°}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{474}\mathit{V}$

(c)0.10MFIJTA at pH 7.00(Find the potential at which$\mathit{C}\mathit{o}\mathbf{\left(}\mathit{H}\mathit{I}\mathbf{\right)}\mathit{T}\mathit{A}{\mathbf{)}}^{\mathbf{2}\mathbf{-}}\mathbf{\mid }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathit{\mu }\mathit{M}$.)

Short answer

1. The cathode potential required to reduce $C{o}^{2+}$to Co is - 0.700V.
2. The cathode potential at which the concentration of cobalt EDTA ion is $10\mu M$is - 0.833V.
3. The cathode potential at which the concentration of cobalt EDTA ion is $10\mu M$at pH seven is - 1.056V.

Step-by-step solution

Concept used.

The voltage of a cell when the electric current is too small,

$\mathit{E}\mathbf{=}\mathit{E}\mathbf{\left(}\mathit{c}\mathit{a}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{e}\mathbf{\right)}\mathbf{-}\mathit{E}\mathbf{\left(}\mathit{a}\mathit{n}\mathit{o}\mathit{d}\mathit{e}\mathbf{\right)}$

E is the electrode's potential that is connected to the current source's negative terminal.

The electrode's potential is E(anode), which is connected to the positive terminal of the current source.

Concentration Polarization: Polarization is defined as a change in product and reactant concentrations at the electrode's surface, although they are the same in solution.

Step 2: The cathode potential required to reduce Co2+ to .

a)

The conversion of cobalt ion into cobalt necessitates a specific cathode potential, which can be calculated as follows:$C{o}^{2+}+2e^{-}\to C{o}_{\left(s\right)}{E}^o=-0.282V$

(cathode Vs S.H.E)$=-0.282-\frac{0.05916}{2}\log \frac{1}{\left|C{o}^{2+}\right|}$

$\begin{array}{r}\left[{\mathrm{Co}}^{2+}\right]=1.0\times {10}^{-6}\mathrm{ME}\\ =-0.459\mathrm{V}\\ E(\text{cathode Vs S.C.E)}\\ =-0.459-0.241\\ =-0.700\mathrm{V}\end{array}$

The cathode potential at which the concentration of cobalt EDTA ion is.

b)

The conversion of the cobalt EDTA ion to cobalt requires a certain cathode voltage, which may be computed as follows:

$Co{(C_2{O}_4{)}_2}^{2-}+2e^{-}\to C{o}_{\left(s\right)}+2C_2{O_4}^{2-}{E}^0=-0.474V$

( cathode Vs S.H.E) $=-0.474V-\frac{0.05916}{2}log\frac{\mid C_2{O_4}^2\cdot T^2}{\mid C{o}^2(C_2{O}_4{)}_2^{2-1}}-0.241$

$\begin{array}{r}\left[{\mathrm{C}}_2{{\mathrm{O}}_4}^{2-}\right]=0.10\mathrm{M}\\ \left[\mathrm{Co}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_2^{2-}\right]=1.0\times {10}^{-6}\mathrm{M}\\ E\text{(cathode Vs S.H.E)}=-0.474\mathrm{V}-\frac{0.05916}{2}\log 100000-0.241\\ E=-0.833\mathrm{V}\end{array}$

The cathode potential at which the concentration of cobalt EDTA ion at pH seven is .

c)

The conversion of the cobalt EDTA ion into cobalt necessitates a certain cathode voltage, which may be computed as follows:

$C{o}^{2+}+2e^{-}\to C{o}_{\left(s\right)}{E}^0=-0.282V$

Formation Constant for$Co(EDTA{)}^{2-}$ is$2.8\times {10}^{16}$

Formation Constant of EDTA is 0.10

$\begin{array}{r}{\alpha }_{y4-}=3.8\times {10}^{-4}\\ \mathrm{Co}\left({\text{EDTA}}^{2-}\right)=1.0\times {10}^{-6}\mathrm{M}\\ \left[{\mathrm{Co}}^{2+}\right]=9.4\times {10}^{-19}\mathrm{M}\\ E=-0.282\mathrm{V}-\frac{0.05916}{2}\log \frac{1}{9.4\times {10}^{19}}-0.241\\ E=-1.056\mathrm{V}\end{array}$