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Q. $1.08 \,g$ 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

Electrochemistry

Solution:

$Ag^+ +\underset{1\,F}{1e^-} \rightarrow \underset{108}{Ag}$
$1 \,F = 1$ mole of electrons $= 96500 \,C$
$0.01 \,F = 1.08 \,g\,Ag$
$\therefore Ag$ left $= 1.08 - 1.08 = 0$