An electric current of 5.00\text{ A} is passed through an aqueous solution of a metal chlo… — JAMB UTME

An electric current of $5.00\text{ A}$ is passed through an aqueous solution of a metal chloride using inert electrodes for $3860\text{ seconds}$ . Calculate the volume of chlorine gas, in $\text{dm}^3$ , liberated at standard temperature and pressure (STP). (Take $1\text{ Faraday} = 96,500\text{ C mol}^{-1}$ , Molar volume of gas at $\text{STP} = 22.4\text{ dm}^3\text{ mol}^{-1}$ )

Answer: 2.24 dm^3

Answer

The volume of chlorine gas liberated at STP is

2.24\text{ dm}^3

Passing

5.00\text{ A}

for

3860\text{ s}

transfers

19,300\text{ C}

of charge, corresponding to

0.200\text{ mol}

of electrons. Because oxidation of chloride ions (

2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-

) requires

2\text{ moles}

of electrons per mole of diatomic chlorine gas,

0.100\text{ mol}

\text{Cl}_2

gas is produced. At STP, this occupies

0.100\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

Step-by-Step Solution

Calculate the total electric charge passed during electrolysis.

Q = 19,300\text{ C}

Using

Q = I \times t

, where current

I = 5.00\text{ A}

and time

t = 3860\text{ s}

Determine the amount of substance of electrons transferred in moles.

n(e^-) = 0.200\text{ mol}

Dividing total charge by Faraday's constant (

96,500\text{ C mol}^{-1}

Apply stoichiometric ratio from the anode half-reaction to find moles of chlorine gas.

n(\text{Cl}_2) = 0.100\text{ mol}

The reaction

2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-

indicates that

2\text{ moles}

of electrons liberate

1\text{ mole}

\text{Cl}_2

Calculate the volume of chlorine gas produced at STP.

V = 2.24\text{ dm}^3

Multiplying the moles of

\text{Cl}_2

by the molar volume of gas at STP (

22.4\text{ dm}^3\text{ mol}^{-1}

Key Concept

Faraday's laws of electrolysis applied to gas volume calculations at STP

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