1.08 g of pure silver was converted into silver nitrate and its solution was taken in a beaker. It was electrolysed using platinum cathode and silver anode. 0.01 faraday of electricity was passed using 0.15 volt above the decomposition potential of silver. The silver content of the beaker after the above shall be

Question

1.08 g of pure silver was converted into silver nitrate and its solution was taken in a beaker. It was electrolysed using platinum cathode and silver anode. 0.01 faraday of electricity was passed using 0.15 volt above the decomposition potential of silver. The silver content of the beaker after the above shall be (A) 0 g (B) 0.108 g (C) 0.18 g (D) 1.08 g

Tardigrade Admin · Accepted Answer

Ag+ + underset1 F1e- arrow underset108Ag 1 F = 1 mole of electrons = 96500 C 0.01 F = 1.08 g Ag ∴ Ag left = 1.08 - 1.08 = 0

0 g

0.108 g

0.18 g

1.08 g

Ag++1F1e−​→108Ag​ 1F=1 mole of electrons =96500C 0.01F=1.08gAg ∴Ag left =1.08−1.08=0

$A g^{+} + 1 F 1 e^{-} \to 108 A g$
$1 F = 1$ mole of electrons $= 96500 C$
$0.01 F = 1.08 g A g$
$∴ A g$ left $= 1.08 - 1.08 = 0$