Question

Chromium metal can be plated out from an acidic solution containing $Cr{{O}_{3}}$ according to the following equation:
$Cr{{O}_{3}}(aq)\text{ + 6}{{H}^{\oplus }}(aq)\text{ + 6}{{e}^{-}}\text{ }\to \text{ }Cr(s)\text{ + 3}{{H}_{2}}O$

The no. of grams of chromium that will be plated out by 2400 C current is:(write your answer to nearest integer)

Answer 1

Hint: In the above chemical equation given above we get to know that there is transfer for 6 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 the electro-chemical equivalent.

Complete step by step answer:
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.

$Cr{{O}_{3}}(aq)\text{ + 6}{{H}^{\oplus }}(aq)\text{ + 6}{{e}^{-}}\text{ }\to \text{ }Cr(s)\text{ + 3}{{H}_{2}}O$
$\begin{align}
  & {{\text{m}}_{{}}}{{\alpha }_{{}}}Q \\
 & {{m}_{{}}}{{=}_{{}}}Z.Q \\
\end{align}$
Or, $m\text{ = }\dfrac{Molar\text{ }mass\text{ x }Q}{n\text{ x }96500}$

Molar mass of Cr = 52g
n-factor (n) = 6
Q = 2400 C
Putting the values, we get:
$m\text{ = }\dfrac{52\text{ x }2400}{6\text{ x }96500}\text{ = 0}\text{.215 g}$

Therefore, the mass of chromium deposited is 0.2 g.

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 amount 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.