we have,
$N=0.1$
volume $(L)=200 \,ml / 1000=0.2 \,L$
no of moles $=$ volume $x$ normality
$=0.2 \times 0.1=0.02$ moles
we know,
$1\, mole$ of $Ag =96500 \,C$
so, $0.02\,moles=0.02 \times 96500=1930 \,C$
we know $q= it ( q=$ charge, $i=c u r r e n t, t=$ time $)$
$1930=0.1 \times t$
$t=19300$ second
now, time taken to remove half the silver from solution $= t / 2=19300 / 2$
$=9650\, seconds$ (Answer)
$F{e^{ + 2 }} + 2{e^ - }\, \to \,Fe\,;\,\,\,\,{E^o} = - 0.440\,V$
$F{e^{ + 3 }} + 3{e^ - }\, \to \,Fe\,;\,\,\,\,{E^o} = - 0.036\,V$
તો $F{e^{ + 3 }} + {e^ - } \to \,F{e^{ + 2 }}$ માટે પ્રમાણિત ઇલેક્ટ્રોન પોટેન્શિયલ $({E^o})$ .............. $\mathrm{V}$ છે.
$(A)$ $Sn^{+4}+ 2e^{-} \rightarrow Sn^{2+}$, $E^o= + 0.15\,V$
$(B)$ $2Hg^{+2} + 2e^{-} \rightarrow Hg_{2}^{+2}$, $E^o = + 0.92\,V$
$(C)$ $PbO_2 + 4H^{+} + 2e^{-} \rightarrow Pb^{+2} + 2H_2O$, $E^o = + 1.45\,V$