Question

A uniform current of 6.0A is required to deposit 78 g of gold from a solution of $AuCl{}_4{}^{-}$. What mass of chlorine gas will be formed simultaneously at the anode of the electrolytic cell?

• A. 35.5 g
• B. 71.0 g
• C. 42.6g
• D. 64.2 g

Answer 1

Answer: (C)

42.6g

The molar mass of $Au$ is 197 g/mol.

$\therefore$ Number of moles of $Au=\frac{mass}{molarmass}=\frac{78}{197}=0.4$

During electrolysis,

$3e^{-}+Au{Cl}_4^{-}\to Au$

$\therefore$ Moles of electron required $=0.4\times 3=1.2$

Charge passed $=1.2\times 96500=115800C$

$\therefore Time=\frac{115800}{6}=19300s$

Now,

1 mole of electron deposit $\frac{1}{2}$mole of $C{l}_2$

$\therefore$ 1.2 mole of electron will deposit 0.6 moles of $C{l}_2$

$\therefore$ Mass of chlorine deposited $=0.6\times 71=42.6g$