Question

During electrolysis of a solution of $AgN{O}_3$, $9650$ coulombs of charge is passed through the solution. What will be the mass of silver deposited on the cathode?

• A. $108$g
• B. $10.8$g
• C. $1.08$g
• D. $216$g

Answer 1

The correct option is B $10.8$g

According to this Faraday's first law, the chemical deposition due to flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.

The given reaction is,

$A{g}^{+}+e^{-}\to Ag$

Amount of Ag deposited by 1F(96500) is 108g

So, amount of Ag deposited by $9650C=\frac{108}{96500}\times 9650=10.8g$