3.5 g of Lead gets deposited when 2 Ampere of current was passed through an aqueous solution of Pb(NO₃)₂. What is the time that was required to deposit this amount of lead?

815.82 seconds

To deposite 207 g Pb, charge required is 96500 C

so, to deposit 3.5 g of Pb,

charge required= 96500 x \(\frac{3.5}{207}\)=1631.64 C

Time for which current passed = \(\frac{Q}{I}\) = \(\frac{1631.64}{207}\) = 815.82 seconds