91影视

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.

a)

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

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

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

b)

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

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

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

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

c)

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

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

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

Formation Constant of EDTA is 0.10

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