Question

How long will it take for a uniform current of $6.00 \,A$ to deposit $78 \,g$ of gold from a solution of $AuCl_4^-$ ? What mass of chlorine gas will be formed simultaneously at anode of the cell?
(Atomic mass of $Au = 197$)

• A. $t = 3010 \,sec$,
  $w = 35.50 \,g$
• B. $t = 20306 \,sec$,
  $w = 45.54 \,g$
• C. $t = 19500 \,sec$,
  $w = 54.5 \,g$
• D. $t = 19139.16 \,sec$,
  $w = 42.24 \,g$

Answer 1

Answer: (D)

Reactions take place at electrodes are :
At cathode : $AuCl^{-}_4 + 3e^- \rightarrow Au +4Cl^-$
At anode : $Cl- \rightarrow \frac{1}{2} Cl_2 +e^-$
For the deposition of $197\, g$ ($1$ mole) of $Au = 3F$ of charge is required thus, for the deposition of $78 \,g$ of $Au$, charge required $=\frac{3}{197}\times78=1.19\,F=1.19\times 96500$ coulombs
From $Q=I\times t$
$t=\frac{Q}{I}=\frac{1.19\times96500}{6}$
$=19139.16\,sec$
By $1 \,F$ charge, $35.5\, g$ of $Cl_2$ gas is formed, thus from $1.19 \,F$, mass of chlorine gas formed $= 35.5 \times 1.19 = 42.24\, g$