Question

Aqueous solution of ZnSO4 is electrolysed by passing current of 965 ampere for 100 s; using inert electrode then total volume of gases liberated at electrode is

Answer 1

Given:

Current, I = 965 ampere

Time, t = 100 s

To find:

The total volume of gases liberated at the electrode

Solution:

Finding the charge through the aqueous solution:

We have, $\boxed{\bold{Q = I\times t}}$

Substituting the given values of I and t in the formula above, we get

The charge "Q" passing through the aqueous solution of ZnSO4 is,

= $965 \times 100$

= $96500 \:C$

Finding the no. of moles of e⁻ :

We have,

$\boxed{\bold{n(e^-) = \frac{Charge}{Faraday's \:constant} }}$

Here

Charge of the given solution = 96500 C

Faraday's constant = 96485 C/mol ≈ 96500 C/mol

∴ $n(e^-) = \frac{96500\:C}{96500\:C/mol} = 1\: mol$

Electrolysis of ZnSO₄:

At Cathode:  $\boxed{\bold{Zn^2^+\:+\:2e^- \Longleftrightarrow Zn}}$

At Anode: $\boxed{\bold{H_2 O \Longleftrightarrow 2H^+\: +\:\frac{1}{2}O_2\:+\:2e^- }}$

From the above reactions, we can say that

On passing 2 moles of 2e⁻, $\frac{1}{2}$ a mole of O₂(g) is evolved

Then, On passing 1 mole of e⁻ = $\frac{\frac{1}{2}}{2}$ = $\frac{1}{4}$ = $0.25 \:mole$ of O₂ is evolved

Finding the volume of gas liberated:

We know ⇒ 1 mole of any gas occupies 22.4 L

∴ 0.25 mole of O₂ will occupy,

= 0.25 mol × 22.4 L

= 5.6 L

Thus, the total volume of gases liberated at the electrode is 5.6 L.

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