QC E Calculate the cathode potential ... [FREE SOLUTION]

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Chapter 17: QC E (page 426)

Calculate the cathode potential (versus S.C.F.) needed to reduce cobalt(II) to $1.0 渭 M$ in each of the following solutions. In each case, Co(s) is the product of the reaction. (Disregard any overpotential.)
(a) $0.10 M H C l O_{4}$
(b) $0.10 M C_{2} O_{4}^{2 -}$ (Find the potential at which) $[C o (C_{2} O_{4})_{2}^{2 -}] = 1.0 渭 M$
role="math" localid="1668354628300" $C o (C_{2} O_{4})_{2}^{2 -} + 2 c^{-} 鈫� C o (s) + 2 C_{2} O_{4}^{2 -} 鈥� 鈥� 鈥� E^{掳} = - 0.474 V$
(c)0.10MFIJTA at pH 7.00(Find the potential at which $C o (H I) T A)^{2 -} 鈭� = 1.0 渭 M$ .)

Short Answer

Expert verified

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

Step by step solution

Concept used.

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

$E = E (c a t h o d e) - E (a n o d e)$

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 .

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

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

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

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

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

$C o {(C_{2} O_{4})_{2}}^{2 -} + 2 e^{-} 鈫� C o_{(s)} + 2 C_{2} {O_{4}}^{2 -} E^{0} = - 0.474 V$

( cathode Vs S.H.E) $= - 0.474 V - \frac{0.05916}{2} l o g \frac{鈭� C_{2} {O_{4}}^{2} 鈰� T^{2}}{鈭� C o^{2} (C_{2} O_{4})_{2}^{2 - 1}} - 0.241$

$\begin{aligned} [C_{2} {O_{4}}^{2 鈭�}] = 0.10 M \\ [Co {(C_{2} O_{4})}_{2}^{2 鈭�}] = 1.0 脳 10^{鈭� 6} M \\ E (cathode Vs S.H.E) = 鈭� 0.474 V 鈭� \frac{0.05916}{2} \log 鈦� 100000 鈭� 0.241 \\ E = 鈭� 0.833 V \end{aligned}$

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

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

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

Formation Constant for $C o (E D T A)^{2 -}$ is $2.8 脳 10^{16}$

Formation Constant of EDTA is 0.10

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