Question

The weight in grams of ${O_2}$ formed at $Pt$ anode during the electrolysis of aqueous ${K_2}S{O_4}$ solution during the passage of one coulomb of electricity is:
A. $\dfrac{{16}}{{96800}}$
B. $\dfrac{8}{{96500}}$
C. $\dfrac{{32}}{{96500}}$
D. $\dfrac{{64}}{{96500}}$

Answer 1

Faraday’s first law of electrolysis states During electrolysis the mass of an ion discharged at the electrode is proportional to the quantity of electricity passed. We know that Faraday is a unit of charge and One Faraday $=$ Charge on one mole of an electron in coulombs $= 96500C/mol$
From this, we can conclude that, Faradays $= \dfrac{Q}{{96500}} = \dfrac{{i \times t}}{{96500}}$
We know that $I = \dfrac{q}{t}$
Where $I$ represents current,
$q$ represents charge,
$t$ represents the time.

Complete step by step answer:
During electrolysis, the reaction takes place as:
${K_2}S{O_4} \rightleftharpoons 2{K^ + } + S{O_4}^{2 - }$ and ${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$
At cathode, hydrogen ions gain two electrons and forms hydrogen gas as:
At cathode the reaction takes place as:
$2{H^ + } + 2{e^ - } \to {H_2}$
At anode, hydroxyl ion forms oxygen gas and water molecules with the loss of four electrons as:
At anode the reaction takes place as:
$4O{H^ - } \to {O_2} + 2{H_2}O + 4{e^ - }$
Here we need to find out the oxygen formed at the Pt electrode.
From above we can see that, 32 gm of ${O_2}$ is obtained from 4 F
Therefore, the weight of oxygen obtained by passing 1C, $\left( {\dfrac{1}{{96500}}} \right) = \dfrac{{32}}{4} \times \dfrac{1}{{96500}} = \dfrac{8}{{96500}}gm$

Therefore, the correct answer is option B.

Note: There are two types of electrodes. One is reactive electrodes which take part in electrode reactions, like copper electrodes and silver electrodes. Another type of electrode is an inert electrode which doesn’t take part in the electrode reaction, like a platinum electrode.
At the cathode, species with high reduction potential is reduced first. Alkali metals and alkaline earth metals are not obtained by electrolysis of their aqueous salt solutions because they have low reduction potential. They are obtained by the electrolysis of their fused salts.
At the anode, species with low reduction potential are oxidized first. Hydroxide ions from water are oxidized at anode because they have lower reduction potential than sulfate ions.
Because of the highest oxidation state of central atom sulfur present in the sulfate ion, so it is not generally oxidized at the anode. If it is to be oxidized at the anode, it is the O-atom which is oxidized.
Thus, products of electrolysis may be different for inert and reactive electrodes and also depend on oxidizing and reducing species present in the electrolytes and state of material being electrolyzed.