Question

Anodic oxidation of ammonium hydrogen sulphate produces ammonium persulphate.
$\begin{align}
& N{{H}_{4}}HS{{O}_{4}}\to N{{H}_{4}}SO_{4}^{-}+{{H}^{+}} \\
& 2N{{H}_{4}}SO_{4}^{-}\to {{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}+2{{e}^{-}}\text{ (anodic oxidation)} \\
& \text{2}{{\text{H}}^{+}}\text{+2}{{\text{e}}^{-}}\to {{\text{H}}_{2}}\text{ (Cathodic reduction)} \\
\end{align}$
Hydrolysis of ammonium persulphate forms ${{H}_{2}}{{O}_{2}}$.
${{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}+2{{H}_{2}}O\to 2N{{H}_{4}}HS{{O}_{4}}+{{H}_{2}}{{O}_{2}}$
Current efficiency in the electrolysis process is $60%$. Calculate the amount of current required to produce $85g$ of ${{H}_{2}}{{O}_{2}}$ per hour.
Hydrolysis reaction shows $100%$ yield.

Answer 1

Hint: It is based on the concept of the Faraday’s first law of electrolysis and the amount of current required can be easily calculated by applying the formula as: $I=\dfrac{w\times 96500}{t\times Z}$ obtained from the Faraday’s law of electrolysis. Now you can easily solve it.

Complete step by step answer:
This is based on Faraday's first law of electrolysis. The Faraday’s first law of electrolysis states that the amount of substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolytic solution.
If w gram of the substance is deposited on passing Q coulombs of electricity, then;
$\begin{align}
& w\text{ }\alpha \text{ }Q \\
& w=ZQ \\
\end{align}$ equation (1)
Here, Z is known as the electrochemical equivalent .
If current of I amperes is passed for t seconds, then;
$Q=I\times t$
Put this value in equation (1) we get;
$w=Z\times I\times t$
Here, electricity is in coulomb , we will convert it into faraday as;
$w=\dfrac{Z\times I\times t}{96500}\text{ (1F=9650C)}$ Equation (2)
Now, considering the statement;
As we have to find the current, so equation (2) changes as;
$I=\dfrac{w\times 96500}{t\times Z}$ Equation (3)
In the statement we have been asked to find the current required to produce $85g$ of ${{H}_{2}}{{O}_{2}}$ , then first we have to find the mass of ${{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}$ required to produce $85g$ of ${{H}_{2}}{{O}_{2}}$.
Now considering the above given reaction ;
${{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}+2{{H}_{2}}O\to 2N{{H}_{4}}HS{{O}_{4}}+{{H}_{2}}{{O}_{2}}$
If $34g$ of ${{H}_{2}}{{O}_{2}}$ is produced by =$228g$ of ${{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}$
Then;
$1g$of ${{H}_{2}}{{O}_{2}}$is produced by =$\dfrac{228}{34}g$ of ${{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}$
And
$85g$of ${{H}_{2}}{{O}_{2}}$is produced by =$\dfrac{228}{34}\times 85g=570g$ of ${{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}$
So, w=$570g$
t= $1\text{ hour =3600 seconds}$ (given)
Equivalent weight i.e. Z is found by dividing the atomic number with the number of electrons required to reduce the cation.
So, will consider the reaction as;
$2N{{H}_{3}}SO_{4}^{-}\to {{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}+2{{e}^{-}}$
Atomic weight of ${{(N{{H}_{4}})}_{2}}{{S}_{2}}{{O}_{8}}$=$228g$
No. of electrons =2
So, $Z=\dfrac{228}{2}=114$
Now , put these all values in equation (3), we get;
$\begin{align}
& I=\dfrac{570\times 96500}{3600\times 114} \\
& \text{ = 134}\text{.02 ampere} \\
\end{align}$
Current efficiency = $60%$
Then the amount of current required=$\dfrac{100}{60}\times 134.02=223.379\text{ ampere}$

So, thus the amount of current required to produce $85g$ of ${{H}_{2}}{{O}_{2}}$ per hour is $223.379\text{ ampere}$.

Note: Always keep in mind to change the units of the species given and always take the standard units. If time is given in seconds, then simply take in seconds and if it is given in hours or minutes, then first convert it into seconds and then apply it in the formula and also for the rest of the units too.