Question

Electrolysis of dilute $H_{2}S{O}_4$ generates $H_2{S}_2{O}_8$ .What current is needed to produce $H_2{S}_2{O}_8$ at rate of 1 mole per hour assuming 75% charge efficiency
a) $71.48$ amp
b) $17.48$ amp
c) $35.74$ amp
d) $53.74$ amp

Answer 1

Hint: The molecular formula $H_{2}S{O}_4$ is used to represent sulphuric acid. It’s a very strong acid and it possesses oxidizing and hygroscopic properties. It is also known as mineral acid.

Complete answer:
The process of passage of an electric current through an electrolyte with subsequent migration of positively and negatively charged ions to the negative and positive electrodes is termed electrolysis. For electrolysis of dilute sulphuric acid, we will be using platinum electrodes.
The chemical equation for electrolysis of sulphuric acid is given as follows,
 $H_{2}S{O}_4\rightleftharpoons H^{+}+HS{O}_4^{-}$
According to the rules of electrolysis the positive ion approaches the cathode and the negative ion approaches the anode
The reaction at cathode proceeds as follows
 $2H^{+}+2e^{-}\to H_2\uparrow$
The reaction at cathode leads to the formation of hydrogen gas
The reaction at anode proceeds as follows
 $2HS{O}_4^{-}\to H_2{S}_2{O}_8+2e^{-}$
Through both the reactions at cathode and anode, we are dealing with two electrons
The amount of current required for one hour can be given as,
 $Q=I\times t$
Assuming that we are doing this at $75\mathrm{\%}$ efficiency for one mole per hour we get,
 $2={\displaystyle \frac{I\times t}{96500}}\times 0.75$
On further solving the equation we get,
 $I={\displaystyle \frac{2\times 96500}{60\times 60\times 0.75}}$
Further simplifying we get,
 $I=71.48$ amp
Hence we conclude that the correct option is a) $71.48$ amp.

Note:
Electrolysis is said to have a wide range of industrial applications. The extraction of metals from their ores, in particular aluminum, and the purification of metals, especially copper, are some of the most important industrial uses. Another large-scale industrial use of electrolysis is in the manufacture of the important alkali, sodium hydroxide. Apart from that, treatment of water which has the presence of heavy metals can also be done with the use of electrolysis.