Question

State Faraday's first law of electrolysis. For the electrode reaction $\text{Z}{{\text{n}}^{\text{2+}}}\text{ + 2}{{\text{e}}^{-}}\text{ }\to \text{ Z}{{\text{n}}_{\text{(s)}}}$, what quantity of electricity in coulombs required to deposit one mole of zinc?

Answer 1

Hint: In the above chemical equation given above we get to know that there is the transfer for 2 moles of electrons for 1 mole of reactant. To find the mass of chromium that will be plated by applying Faraday’s first law of electrolysis. The formula for the law is given below:
$\begin{align}
& m\text{ }\alpha \text{ }Q \\
& m\text{ = }Z.Q \\
\end{align}$
where,
m is the mass of electrolyte deposited,
Q is the quantity of electricity deposited,
Z is the constant of proportionality and is known as electrochemical equivalent.

Complete step-by-step solution:
Faraday’s First Law of Electrolysis states that the chemical deposition of a substance due to the flow of current through an electrolyte is directly proportional to the quantity of electricity passed through it.
$\text{Z}{{\text{n}}^{\text{2+}}}\text{ + 2}{{\text{e}}^{-}}\text{ }\to \text{ Z}{{\text{n}}_{\text{(s)}}}$
For the above electrode reaction, we see that there is a gain of 2 electrons for 1 zinc ion. So, for 1 mole of zinc ion, the number of electrons involved is 2 moles.
We know that 1 F or 1 farad is the amount charge present in 1 mole of electrons. So, the total charge in the above reaction is 2F.
1F = 96500 C
2F = 193000 C

Therefore, the quantity of electricity required to deposit one mole of zinc is 193000 coulombs.

Note: Faraday had given two laws on electrolysis. The first law is discussed above and the second law is given below:
Faraday’s second law states that the number of substances deposited due to passage of the same amount of electric current will be proportional to their respective equivalent weights.
Equivalent weight is the molar mass of the substance divided by the n-factor of the substance.