Question

Suppose that gold is being plated onto another metal in an electrolytic cell. The half-cell reaction producing the Au(s) is $AuC{l}_4^{\ominus }\to Au\left(s\right)+4C{l}^{\ominus }+3e^{-}$
If a $0.30$ $-A$ current runs for $1.50$ min, what mass of $Au\left(s\right)$ will be plated, assuming all the electrons are used in the reduction of $AuC{l}_4^{\ominus }$?

• A. $0.184g$
• B. $0.551g$
• C. $1.184g$
• D. $0.613g$

Answer 1

The correct option is A $0.184g$

Number of Faradays = $\frac{It}{96500}=\frac{0.5\times 15\times 60}{96500}=2.8\times {10}^{-3}$
$AuC{l}_4^{\ominus }\to Au\left(s\right)+C{l}^{\ominus }+3e^{-}$
$3F$ $\equiv$ 1 mol $Au$
$2.8\times {10}^{-3}F\equiv \frac{2.8}{3}\times {10}^{-3}molAu$
$=\frac{2.8}{3}\times 197\times {10}^{-3}gAu$
$=0.184g$ $Au$