Suppose that gold is being plated on to another metal in an electrolytic cell. The half-cell reaction producing the Au(.s.) is AuCl4-+3e- arrow Au(s)+4(Cl)- . If a 0.30A current passed for 15min , what mass of Au(.s.) will be plated, assume all the electrons are used in the reduction of AuCl4- ? The Faraday constant is 96485C/mol and molar mass of Au textis197 .

Question

Suppose that gold is being plated on to another metal in an electrolytic cell. The half-cell reaction producing the Au(.s.) is AuCl4-+3e- arrow Au(s)+4(Cl)- . If a 0.30A current passed for 15min , what mass of Au(.s.) will be plated, assume all the electrons are used in the reduction of AuCl4- ? The Faraday constant is 96485C/mol and molar mass of Au textis197 . (A) 0.184g (B) 0.551g (C) 1.84g (D) 0.613g

Tardigrade Admin · Accepted Answer

Mass of Au deposited = Number of Faraday passed × Eq. mass =(0.30 × 15 × 60/96485)× (197/3)=0.184 textg

$0.184 g$

$0.551 g$

$1.84 g$

$0.613 g$

Mass of Au deposited = Number of Faraday passed $\times$ Eq. mass
$= \frac{0.30 \times 15 \times 60}{96485} \times \frac{197}{3} = 0.184 g$

0.184g

0.551g

1.84g

0.613g

Mass of Au deposited = Number of Faraday passed × Eq. mass =964850.30×15×60​×3197​=0.184g

$0.184 g$

$0.551 g$

$1.84 g$

$0.613 g$

Mass of Au deposited = Number of Faraday passed $\times$ Eq. mass
$= \frac{0.30 \times 15 \times 60}{96485} \times \frac{197}{3} = 0.184 g$