Question

If $0.5L$ of a $0.6MSnS{O}_4$ solution is electrolysed for a period of $30$ min, using a current of $4.6A$. If insert electrodes areas. What is the final concentration of $S{n}^{2+}$ remaining in the solution?

Answer 1

For the deeposition of $Sn$ electrochemical reaction is

${Sn}^{2+}+2e^{-}\to Sn$

Using Faraday's first law of electrolysis: Amount of substance deposited (m) at an electrode is directly proportional to amount of charge passed through the solution and is written as

$m=\frac{M}{n\times 96500}It$

where $M=118.7g/mol,n=2,I=4.6A,t=30/times60s$

Amount of $Sn$ deposited, $m=\frac{118.7}{2\times 96500}4.6\times 30\times 60$

$m=5.0924g=85.8mM$ from 0.5 L

Amount remaining in solution:

$0.6-0.0858=0.514M$