Question

Calculate the time required for a constant current of 0.875 A to deposit 0.350 g of (a) Tl(III) as the element on a cathode, (b) Tl(I) as the Tl2O3 on an anode, and (c) Tl(I) as the element on a cathode.

Short answer

(a) The time required for constant current to deposit Tl (III) as the element on a cathode is 9.44 min.

(b) The time required for constant current to deposit Tl(I) as the Tl₂O₃ on an anode is 6.3 min.

(c) The time required for constant current to deposit Tl(I) as the element on a cathode is 3.15 min.

Step-by-step solution

Part (a) Step 1: Given Information

The time required for constant current to deposit Tl(III) as the element on a cathode should be determined.

Part (a) Step 2: Explanation

The weight of the element Tl(III) is 0.350 g and the constant current is 0.875 A.

The molecular mass of Tl(III) is 204.37 g.

The expression for the Faraday’s equation is:

$Q=n_{A}nF$…… (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:

$Q=It$…… (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.

data-custom-editor="chemistry" ${\mathrm{Tl}}^{3+}+3{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Tl}\left(\mathrm{s}\right)$

The number of electrons gained by cobalt is 3. Thus, n=3.

Substitute 0.350 g for mass of element and 204.37 g for molar mass in Equation (III).

$$\begin{gathered} n_A=\left(\frac{0.350g}{204.37g}\right) \\ =0.001713 \end{gathered}$$

Substitute 0.001713 for n_A, 3 for n, and 96485 for F in Equation (I).

$$\begin{gathered} Q=(0.001713)3(96485C) \\ =495.72C \end{gathered}$$

Substitute 495.72 C for Q and 0.875 A in Equation (II).

$$\begin{gathered} 95.72\mathrm{C}=(0.845\mathrm{A})\mathrm{t} \\ \mathrm{t}=\frac{495.72\mathrm{C}}{(0.875\mathrm{A})\left({\displaystyle \frac{1\mathrm{C}/\mathrm{s}}{1\mathrm{A}}}\right)} \\ \mathrm{t}=(566.53\mathrm{s})\left(\frac{1\min }{60\mathrm{s}}\right) \\ \mathrm{t}\approx 9.44\min \end{gathered}$$

Therefore, the time required for constant current to deposit Tl(III) as the element on a cathode is 9.44 min.

Part (b) Step 1: Given Information

The time required for constant current to deposit Tl(I) as the Tl₂O₃ on an anode should be determined.

Part (b) Step 2: Explanation

The weight of the element Tl(I) is 0.350 g and the constant current is 0.875 A.

The molecular mass of Tl(I) is 204.37 g.

The reaction is given below.

$2{\mathrm{Tl}}^{+}+3{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{Tl}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+6{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$

The number of electrons gained 2 moles of cobalt is 4. Thus, n=2.

Substitute 0.350 g for mass of element and 204.37 g for molar mass in Equation (III).

$$\begin{gathered} n_A=\left(\frac{0.350g}{204.37g}\right) \\ =0.001713 \end{gathered}$$

Substitute 0.001713 for n_A, 2 for n, and 96485 for F in Equation (I).

$$\begin{gathered} Q=(0.001713)2(96485C) \\ =330.48C \end{gathered}$$

Substitute 330.48 C for Q and 0.875 A in Equation (II).

localid="1649392373694" $$\begin{gathered} 330.48\mathrm{C}=(0.875\mathrm{A})\mathrm{t} \\ \mathrm{t}=\frac{330.48\mathrm{C}}{(0.875\mathrm{A})\left({\displaystyle \frac{1\mathrm{C}/\mathrm{s}}{1\mathrm{A}}}\right)} \\ \mathrm{t}=(377.69\mathrm{s})\left(\frac{1\min }{60\mathrm{s}}\right) \\ \mathrm{t}\approx 6.3\min \end{gathered}$$

Therefore, the time required for constant current to deposit Tl(I) as the Tl₂O₃ on an anode is 6.3 min.

Part (c) Step 1: Given Information

The time required for constant current to deposit Tl(I) as the element on a cathode should be determined.

Part (c) Step 2: Explanation

The reaction is given below.

${\mathrm{Tl}}^{+}+1{\mathrm{e}}^{-}\rightleftharpoons \mathrm{Tl}\left(\mathrm{s}\right)$

The number of electrons gained by cobalt is 1. Thus, n=1.

Substitute 0.350g for mass of element and 204.37 g for molar mass in Equation (III).

$$\begin{gathered} n_A=\left(\frac{0.350g}{204.37g}\right) \\ =0.001713 \end{gathered}$$

Substitute 0.001713 for n_A, 1 for n, and 96485 for F in Equation (I).

$$\begin{gathered} Q=(0.001713)1(96485C) \\ =165.279C \end{gathered}$$

Substitute 165.279 C for Q and 0.875 in Equation (II).

$$\begin{gathered} 165.279\mathrm{C}=(0.875\mathrm{A})\mathrm{t} \\ \mathrm{t}=\frac{165.279\mathrm{C}}{(0.875\mathrm{A})\left({\displaystyle \frac{1\mathrm{C}/\mathrm{s}}{1\mathrm{A}}}\right)} \\ \mathrm{t}=(188.89\mathrm{s})\left(\frac{1\min }{60\mathrm{s}}\right) \\ \mathrm{t}\approx 3.15\min \end{gathered}$$

Therefore, the time required for constant current to deposit Tl(I) as the element on a cathode is 3.15 min.