Suppose that gold is being plated on to another metal in an electrolytic cell. The half-cell reaction producing the Au(s) is A u C l 4 - +3e - → A u s + 4 C l - . If a 0.30 A current runs for 15.00 mi

Question

Suppose that gold is being plated on to another metal in an electrolytic cell. The half-cell reaction producing the Au(s) is $A u C l_{4}^{-} {+3e}^{-} \to A u (s) + 4 C l^{-}$ . If a 0.30 A current runs for 15.00 minute, what mass of Au(s) will be plated, assume all the electrons are used in the reduction of $A u C l_{4}^{-}$ ? The Faraday constant is 96485 $C / m o l$ and molar mass of Au is 197.

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Answer:

A

Solution:

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

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

TestHub · Accepted Answer

Mass of Au deposited = Number of Faraday passed × Eq. mass = 0.30 × 15 × 60 96485 × 197 3 = 0.184 g

Chemistry - Electrochemistry Question with Solution | TestHub

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