What time, in seconds, is required to deposit 4.5\text{ g} of aluminium at the cathode dur… — JAMB UTME

What time, in seconds, is required to deposit $4.5\text{ g}$ of aluminium at the cathode during the electrolysis of molten aluminium oxide using a constant current of $5.0\text{ A}$ ? $[\text{Al} = 27,\, 1\text{ F} = 96,500\text{ C mol}^{-1}]$

$9650\text{ s}$ Answer
B
$3217\text{ s}$
C
$6433\text{ s}$
D
$86850\text{ s}$

Answer

The time required to deposit

4.5\text{ g}

of aluminium is

9650\text{ s}

The deposit of

4.5\text{ g}

\text{Al}

corresponds to

0.1667\text{ mol}

. Since

\text{Al}^{3+}

requires 3 electrons per atom deposited (

z = 3

), the total charge needed is

0.5\text{ F} = 48,250\text{ C}

. Dividing this charge by a current of

5.0\text{ A}

gives

9650\text{ seconds}

Step-by-Step Solution

Determine the moles of aluminium deposited and the moles of electrons transferred.

Moles of

\text{Al} = \frac{4.5\text{ g}}{27\text{ g mol}^{-1}} = 0.1667\text{ mol}

. Reduction half-reaction:

\text{Al}^{3+} + 3e^- \rightarrow \text{Al}

. Therefore,

z = 3

moles of electrons are required per mole of

\text{Al}

Aluminium is a trivalent metal (

z = 3

), so depositing 1 mole of

\text{Al}

requires 3 Faradays of charge.

Calculate the total quantity of electricity (

Q

) required in coulombs.

Q = n_{e^-} \times F = (0.1667\text{ mol} \times 3) \times 96,500\text{ C mol}^{-1} = 0.5 \times 96,500 = 48,250\text{ C}

Total charge is the product of moles of electrons transferred and Faraday's constant.

Calculate the required time (

t

) using

Q = I \times t

t = \frac{Q}{I} = \frac{48,250\text{ C}}{5.0\text{ A}} = 9650\text{ s}

Dividing total charge by current yields time in seconds.

Key Concept

Quantitative application of Faraday's first and second laws of electrolysis for multivalent ions.

Estimated Time:1m 30s

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