Question

An aqueous solution of NaCl on electrolysis gives H_2(g), Cl_2(g) and NaOH according to the reaction

\[\ce{2Cl^-_{ (aq)} + 2H2O -> 2OH^-_{ (aq)} + H2_{(g)} + Cl2_{(g)}}\].

A direct current of 25 amperes with a current efficiency of 62% is passed through 20 litres of NaCl solution (20% by weight).

1. Write down the reactions taking place at the electrodes.
2. How long will it take to produce 1 kg of Cl₂?
3. What will be the molarity of solution with respect to OH⁻? Assume no loss in volume due to evaporation.

Answer 1

Given: The given reaction is,

\[\ce{2Cl^-_{ (aq)} + 2H2O -> 2OH^-_{ (aq)} + H2_{(g)} + Cl2_{(g)}}\]

Current = 25 A

Current efficiency = 62%

Volume of solution = 20 L

20% NaCl by weight

Molar mass of Cl₂ = 71 g/mol

To find:

1. Electrode reactions = ?
2. Time to produce 1 kg Cl₂ = ?
3. Molarity of OH⁻ formed = ?

Calculation:

i. Electrode reactions:

At Anode (oxidation):

\[\ce{2Cl- -> Cl2 + 2e-}\]

At Cathode (reduction):

\[\ce{2H2O + 2e- -> H2 + 2OH-}\]

ii. Time to produce 1 kg of Cl₂:

Moles of Cl₂ = `("Mass of Cl"_2)/("Molar mass of Cl"_2)`

= `1000/71`

= 14.0845 mol

From the half-reaction:

\[\ce{2Cl- -> Cl2 + 2e-}\] ⇒ 2 mol e⁻ per mol Cl_2​

Total moles of electrons = 2 × 14.0845

= 28.169 mol e⁻

Charge (Q) = 28.169 × 96500

= 2.718 × 10⁶ C

Effective charge (t) = `(4.383 xx 10^6)/25`

= 1.7532 × 10⁵ seconds

= `(1.7532 xx10^5)/3600`

= 48.71 hours

iii. Molarity of OH⁻ in the solution:

From the reaction:

1 mol of Cl₂ gives 2 mol of OH⁻.

We know that,

Moles of Cl_2​ = 14.0845

⇒ Moles of OH⁻ = 2 × 14.0845 = 28.169 mol

Molarity of OH⁻ = `("Moles of OH"^-)/"Volume of solution"`

= `28.169/20`

= 1.408 M