A current of 2.0\text{ A} is passed through an aqueous solution of copper(II) tetraoxosulf… — JAMB UTME

A current of $2.0\text{ A}$ is passed through an aqueous solution of copper(II) tetraoxosulfate(VI) for $965\text{ seconds}$ . What is the mass of copper deposited at the cathode? $[F = 96,500\text{ C mol}^{-1},\text{ Cu} = 64]$

$0.64\text{ g}$ Answer
B
$1.28\text{ g}$
C
$2.56\text{ g}$
D
$0.32\text{ g}$

Answer

The mass of copper deposited at the cathode is

0.64\text{ g}

The total charge passed is

Q = 2.0\text{ A} \times 965\text{ s} = 1930\text{ C}

. Dividing by Faraday's constant yields

0.02\text{ mol}

of electrons. Because copper(II) ions require

2\text{ moles}

of electrons per mole of copper metal (

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

), the amount of copper deposited is

0.01\text{ mol}

, which corresponds to

0.01 \times 64 = 0.64\text{ g}

Step-by-Step Solution

Calculate total quantity of electricity (

Q

) passed

Q = I \times t = 2.0\text{ A} \times 965\text{ s} = 1930\text{ C}

Faraday's first law relates charge to current and time.

Calculate the moles of electrons transferred

\text{Moles of } e^- = \frac{Q}{F} = \frac{1930\text{ C}}{96,500\text{ C mol}^{-1}} = 0.02\text{ mol } e^-

One mole of electrons carries one Faraday (

96,500\text{ C}

) of charge.

Determine moles of copper deposited using electrode reduction half-equation

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \implies n(\text{Cu}) = \frac{0.02\text{ mol } e^-}{2} = 0.01\text{ mol}

Reduction of one

\text{Cu}^{2+}

ion requires

2

moles of electrons per mole of copper deposited.

Calculate mass of copper deposited

\text{Mass} = n \times \text{Molar Mass} = 0.01\text{ mol} \times 64\text{ g mol}^{-1} = 0.64\text{ g}

Multiplying moles of substance by its molar mass yields mass.

Key Concept

Faraday's First Law of Electrolysis and Quantitative Stoichiometry of Charge

Estimated Time:1m 0s

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