A steady electric current is passed through an aqueous solution of zinc tetraoxosulfate(VI… — JAMB UTME

A steady electric current is passed through an aqueous solution of zinc tetraoxosulfate(VI) for $4825\text{ s}$ . If $3.25\text{ g}$ of zinc is deposited at the cathode, what is the magnitude of the electric current, in Amperes, used?

$[\text{Zn} = 65,\text{ 1 F} = 96500\text{ C mol}^{-1}]$

Answer: 2 A

Answer

The magnitude of the electric current required is

2.0\text{ A}

Depositing

3.25\text{ g}

\text{Zn}

(atomic mass

65\text{ g mol}^{-1}

) requires

0.05\text{ mol}

of zinc metal. Since each

\text{Zn}^{2+}

ion requires

2

electrons to be reduced,

0.10\text{ mol}

of electrons (

9650\text{ C}

) must pass through the electrolyte. Dividing this charge by time (

4825\text{ s}

) gives

2.0\text{ A}

Step-by-Step Solution

Calculate the moles of zinc deposited at the cathode.

Moles of

\text{Zn} = \frac{3.25\text{ g}}{65\text{ g mol}^{-1}} = 0.05\text{ mol}

Dividing the mass of metal deposited by its relative atomic mass gives the number of moles deposited.

Determine the quantity of electricity in Coulombs needed for the deposition.

Reduction half-reaction:

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

. Moles of

e^- = 2 \times 0.05\text{ mol} = 0.10\text{ mol}

. Quantity of electricity

Q = 0.10\text{ mol} \times 96500\text{ C mol}^{-1} = 9650\text{ C}

Faraday's second law relates the mole ratio of electrons to metal ion charge.

Calculate the steady electric current in Amperes.

I = \frac{Q}{t} = \frac{9650\text{ C}}{4825\text{ s}} = 2.0\text{ A}

Electric current is defined as the rate of charge flow over time (

I = \frac{Q}{t}

Key Concept

Faraday's Laws of Electrolysis and Quantitative Calculations

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