Question

A factory produces 40 kg of calcium in two hours by electrolysis. How much aluminium can be produced by the same current in two hours? (Ca = 40 u, Al = 27 u) [HINT: Same qnt. of electricity is passed]

• A. 22 kg
• B. 18 kg
• C. 9 kg
• D. 27 kg

Answer 1

Answer: (B)

18 kg

Answer 2

According to Faraday's second law of electrolysis, when the same quantity of electricity is passed through different electrolytes, the masses of the substances deposited or liberated at the electrodes are directly proportional to their equivalent weights.

The equivalent weight ($E$) of a substance is given by its atomic mass ($M$) divided by its valency ($v$).

$E = \frac{M}{v}$

For calcium (Ca), the atomic mass $M_{Ca} = 40$ u. In electrolysis producing calcium metal, the ion is $\text{Ca}^{2+}$, so the valency $v_{Ca} = 2$. The equivalent weight of calcium is $E_{Ca} = \frac{M_{Ca}}{v_{Ca}} = \frac{40}{2} = 20$.

For aluminium (Al), the atomic mass $M_{Al} = 27$ u. In electrolysis producing aluminium metal, the ion is $\text{Al}^{3+}$, so the valency $v_{Al} = 3$. The equivalent weight of aluminium is $E_{Al} = \frac{M_{Al}}{v_{Al}} = \frac{27}{3} = 9$.

We are given that 40 kg of calcium is produced ($W_{Ca} = 40$ kg). The same current is passed for the same time (2 hours), which means the same quantity of electricity is used for both calcium and aluminium production.

Applying Faraday's second law:

$\frac{W_{Ca}}{E_{Ca}} = \frac{W_{Al}}{E_{Al}}$

Substitute the known values:

$\frac{40}{20} = \frac{W_{Al}}{9}$

Simplify the left side:

$2 = \frac{W_{Al}}{9}$

Solve for $W_{Al}$:

$W_{Al} = 2 \times 9 = 18$ kg

Therefore, 18 kg of aluminium can be produced by the same current in two hours.