Question

Suppose that gold is being plated on to another metal in an electrolytic cell. The half-call reaction producing :
$AuC{l}_4{}^{-}\to Au\left(s\right)+4C{l}^{-}-3e^{-}$

If a 0.30 $A$ current runs for 15.00 min, w mass of $Au\left(s\right)$ will be plated, assuming all electrons are used in the reduction of $AuC{l}_4{}^{-}$. The Faraday constant is 96485 $C/mol$. Molar mass of $Au$ is 197.

• A. 0.184 g Au
• B. 0.551 g Au
• C. 1. 84 g Au
• D. 0. 613 g Au

Answer 1

Answer: (A)

0.184 g Au

Mass of Au deposited $=$ Number of Faraday passed $\times$ Eq. mass

$\frac{0.30\times 15\times 60}{96485}\times \frac{197}{3}=0.184g$