Question

What is the time (in sec) required for deposition all the silver present in 125ml of 1M $AgN{O}_3$ solution by passing a current of 241.25 amperes?

( $1F=96500$ coulombs)

• A. 10
• B. 50
• C. 1000
• D. 100

Answer 1

The correct option is B 50

$\text{Moles of}AgN{O}_3=M\times V$

$=1\times \frac{125}{1000}=0.125$ mole

$A{g}^{+}+e^{-}\to Ag$
1 mole $Ag$ required 1 mole $e^{-}$

$\therefore 0.125$ mole $Ag$ required 0.125 mole $e^{-}$

$\therefore \text{Total Charge}=0.125\times 96500$

$\therefore 0.125\times 96500=t\times 241.25$

$\therefore t=50sec$

Hence, option B is correct.