Question

Calculate the time required for a constant current of 0.800 A to deposit 0.250 g of (a) Co(II) as the element on a cathode and (b) as Co₃O₄ on an anode. Assume 100% current efficiency for both cases.

Short answer

(a) The time required for constant current to deposit Co(II) as the element on a cathode is 17 min.

(b) The time required for constant current to deposit Co₃O₄ as the element on an anode is 5.68 min.

Step-by-step solution

Part (a) Step 1: Given Information

The weight of the element Co(II) is 0.250 g and the constant current is 0.800 A.

Part (a) Step 2: Explanation

The molecular mass of Co(II) is 58.93 g.

The expression for the Faraday’s equation is:

$\mathrm{Q}={\mathrm{n}}_{\mathrm{A}}\mathrm{nF}$…… (I)

Here, the charge is Q, the number of moles in analyte is n_A, the number of electrons is n, and the Faraday’s constant is F.

The expression for the quantity of charge is:

$\mathrm{Q}=\mathrm{It}$…… (II)

Here, the constant current is I and the time taken is t.

The expression for the number of moles of an element is:

$\mathrm{n}=\frac{\mathrm{mass}\mathrm{of}\mathrm{element}}{\mathrm{molar}\mathrm{mass}}$ …… (III)

The reaction is given below.

${\mathrm{Co}}^{2+}+2{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Co}\left(\mathrm{s}\right)$

The number of electrons gained by cobalt is 2. Thus, n=2.

Substitute 0.25 g for mass of element and 58.93 g for molar mass in Equation (III).

$$\begin{gathered} n_A=\left(\frac{0.25\mathrm{g}}{58.93\mathrm{g}}\right) \\ =0.00424 \end{gathered}$$

Substitute 0.00424 for n_A, 2 for n, and 96485 for F in Equation (I).

localid="1649312206821" $$\begin{gathered} Q=\left(0.00424\right)2(96485C) \\ =818.193C \end{gathered}$$

Substitute 818.193 C for Q and 0.800 A in Equation (II).

$$\begin{gathered} 818.193\mathrm{C}=(0.800\mathrm{A})\mathrm{t} \\ \mathrm{t}=\frac{818.193\mathrm{C}}{(0.800\mathrm{A})\left({\displaystyle \frac{1\mathrm{C}/\mathrm{s}}{1\mathrm{A}}}\right)} \\ \mathrm{t}=(1022.714)\left(\frac{1\min }{60\mathrm{s}}\right) \\ \mathrm{t}\approx 17\min \end{gathered}$$

Therefore, the time required for constant current to deposit Co(II) as the element on a cathode is 17 min.

Part (b) Step 1: Given Information

The weight of the element Co₃O₄ is 0.250 g and the constant current is 0.800 A.

Part (b) Step 2: Explanation

The molecular mass of cobalt is 58.93 g.

The expression for the Faraday’s equation is:

$\mathrm{Q}={\mathrm{n}}_{\mathrm{A}}\mathrm{nF}$ …… (I)

Here, the charge is Q, the number of moles in analyte is n_A, the number of electrons is n, and the Faraday’s constant is F.

The expression for the quantity of charge is:

$\mathrm{Q}=\mathrm{It}$…… (II)

Here, the constant current is I and the time taken is t.

The expression for the number of moles of an element is:

$\mathrm{n}=\frac{\mathrm{mass}\mathrm{of}\mathrm{element}}{\mathrm{molar}\mathrm{mass}}$…… (III)

The reaction is given below.

$3{\mathrm{Co}}^{2+}+4{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{Co}}_3{\mathrm{O}}_4\left(\mathrm{s}\right)$

The number of electrons gained 3 moles of cobalt is 2. Thus, n= 2/3 .

Substitute 0.25 g for mass of element and 58.93 g for molar mass in Equation (III).

$$\begin{gathered} n_A=\left(\frac{0.25\mathrm{g}}{58.93\mathrm{g}}\right) \\ =0.00424 \end{gathered}$$

Substitute 0.00424 for n_A, 2/3 for n, and 96485 C for F in Equation (I).

$$\begin{gathered} Q=\left(0.00424\right)\frac{2}{3}(96485C) \\ =272.73C \end{gathered}$$

Substitute 272.73 C for Q and 0.800 A in Equation (II).

$$\begin{gathered} 272.73\mathrm{C}=(0.800\mathrm{A})\mathrm{t} \\ \mathrm{t}=\frac{272.73\mathrm{C}}{(0.800\mathrm{A})\left({\displaystyle \frac{1\mathrm{C}/\mathrm{s}}{1\mathrm{A}}}\right)} \\ \mathrm{t}=(340.91s)\left(\frac{1\min }{60\mathrm{s}}\right) \\ \mathrm{t}\approx 5.68\min \end{gathered}$$

Therefore, the time required for constant current to deposit Co₃O₄ as the element on an anode is 5.68 min.