Question

How much time will it take for a uniform current of 6.00A to deposit 78g of gold from a solution of $AuCl_{4}^{-}$.​? What mass of chlorine gas will be formed simultaneously at the anode of the cell? (Atomic mass of Au =197)

Answer 1

Hint: Find the moles of Au needed to deposit 78g of Gold. Now multiply it by a change in oxidation number change in the reaction. The amount of charge to be transferred will be (moles of A) x (change in oxidation number). Divide it by the value of current and you will have the time. Write the reaction and see what amount of $Cl_2$ released for a given amount of Au deposited

Complete step by step solution:
Calculation of oxidation number of Au in $AuCl_{4}^{-}$
Let oxidation number of Au be x
$\implies$ x -4(For each Cl) = -1
$\implies$ x = +3
$\implies$ The oxidation number of Au is +3
When Au is deposited it gets reduced to 0, so change in oxidation number of Au is 3
$\implies$ the number of electrons needed to reduce(deposit) one atom of Au is 3
Moles of Au required to deposit 78g of Au = $\dfrac{m}{M}$
Where m is mass of Au to be deposited
M is molar mass of Au
$\implies n $ n = $\dfrac{78}{197}$
The amount of current required = n x 3 x $N_A$ x e = I t
$N_A$ is the Avogadro number
e is charge on electron
Also $N_A$ x e = F
F is Faraday’s Constant whose value is approximately equal to 96485 C/mol
I is the current, t is time
Putting all known values gives t=$\dfrac{3 \times 78 \times 96485}{197 \times 6}$
$\implies $ t= 19101 sec.
$Cl^{-1}$ in the given compound will reduce to $Cl_2$
One mole of 2F​ will give one mole of $Cl_2$
$\implies $ mass of $Cl_2$ liberated = $\dfrac{n \times 3 \times 71}{2}$ $\dfrac{n \times 3 \times 71}{2}$
 71g is Molar mass of $Cl_2$
. $\implies $ Mass of chlorine gas liberated = 42.16 g

So, for the given options option D is the closest hence it is the answer.

Note: All of the Cl from the given compound will not be liberated as the number of electrons required for all 4 chlorine atoms to reduce is not sufficient. Also, note that the above formula used is the mathematical representation of Faraday’s first law of electrolysis.