$1.08 \mathrm{~g}$ of pure silver was converted into silver nitrate and its solution was taken in a beaker.…
$1.08 \mathrm{~g}$ 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
$0 \mathrm{~g}$
$0.108 \mathrm{~g}$
$1.08 \mathrm{~g}$
None of these
Solution
$\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}$
$1 \mathrm{~F}=1$ mole of electrons $=96500 \mathrm{C}$
$0.01 \mathrm{~F}=1.08 \mathrm{~g} \mathrm{Ag} ; \mathrm{Ag}$ left $=1.08-1.08=0 \mathrm{~g}$
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