During the electrolytic refining of copper, a steady current of 5.0\text{ A} is passed thr… — JAMB UTME

During the electrolytic refining of copper, a steady current of $5.0\text{ A}$ is passed through an aqueous copper(II) tetraoxosulfate(VI) ( $CuSO_4$ ) solution for $965\text{ seconds}$ . What mass of copper, in grams, is deposited at the cathode? (Faraday's constant $F = 96,500\text{ C mol}^{-1}$ ; Molar mass of $Cu = 64\text{ g mol}^{-1}$ )

Answer: 1.6 g

Answer

1.6 g

Passing a

5.0\text{ A}

current for

965\text{ s}

transfers

4825\text{ C}

of electric charge, corresponding to

0.05\text{ mol}

of electrons. Because the deposition of copper from

CuSO_4

follows

Cu^{2+} + 2e^- \rightarrow Cu

, every

2\text{ moles}

of electrons deposit

1\text{ mole}

of copper metal. Thus,

0.025\text{ mol}

of copper is deposited, which corresponds to

0.025\text{ mol} \times 64\text{ g mol}^{-1} = 1.6\text{ g}

Step-by-Step Solution

Calculate the total quantity of electricity (

Q

) passed through the electrolyte.

Q = I \times t = 5.0\text{ A} \times 965\text{ s} = 4825\text{ C}

Electric charge is the product of current in amperes and time in seconds.

Calculate the moles of electrons transferred.

n(e^-) = \frac{Q}{F} = \frac{4825\text{ C}}{96500\text{ C mol}^{-1}} = 0.05\text{ mol}

of electrons

One mole of electrons carries a charge equivalent to 1 Faraday (

96,500\text{ C}

Use the cathode half-equation to find the moles of deposited copper.

Cathode reaction:

Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}

. Moles of

Cu = \frac{0.05\text{ mol}}{2} = 0.025\text{ mol}

Reduction of one mole of copper(II) ions requires two moles of electrons.

Convert the moles of deposited copper into mass.

Mass of

Cu = n \times M = 0.025\text{ mol} \times 64\text{ g mol}^{-1} = 1.6\text{ g}

Multiplying the chemical amount of copper by its molar mass yields the mass in grams.

Key Concept

Quantitative electrolysis of copper(II) ions using Faraday's laws of electrolysis

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