Question

What is the time required (in seconds) for depositing the silver present in $125ml$ of $1MAgN{O}_3$ solution passing a current of $241.25A[1F=96500C]?$

• A. 10 sec
• B. 50 sec
• C. 1000 sec
• D. 100 sec

Answer 1

Answer: (B)

$1000ml$ of $AgN{O}_3$ solution contains $108gAg$

$\therefore 125ml$ of $AgN{O}_3$ solution will contain

$=\frac{108\times 125}{1000}=13.5gAg$

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

(1 F =96500) 108 g

$108g$ of $Ag$ is deposited by $96500C$

$\therefore 13.5$ of $Ag$ is deposited $by=\frac{96500}{108}\times 13.5=12062.5C$

$Q=It$

$t=\frac{Q}{I}=\frac{12062.5}{241.25}=50$ sec