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(B) The quantity of electricity required to deposit $1 \ mol$ of a metal ion $M^{n+}$ is given by $n 	imes F$,where $F = 96500 \ C \ mol^{-1}$.$a. \

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$a-iii, \ b-iv, \ c-ii, \ d-i$

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(B) The quantity of electricity required to deposit $1 \ mol$ of a metal ion $M^{n+}$ is given by $n 	imes F$,where $F = 96500 \ C \ mol^{-1}$.$a. \ Ag^{+} + e^{-} ightarrow Ag$: $n = 1$,so $1 	imes 96500 = 96500 \ C \ mol^{-1}$ $(iii)$.$b. \ Mg^{2+} + 2e^{-} ightarrow Mg$: $n = 2$,so $2 	imes 96500 = 193000 \ C \ mol^{-1}$ $(iv)$.$c. \ Al^{3+} + 3e^{-} ightarrow Al$: $n = 3$,so $3 	imes 96500 = 289500 \ C \ mol^{-1}$ $(ii)$.$d. \ Ti^{4+} + 4e^{-} ightarrow Ti$: $n = 4$,so $4 	imes 96500 = 386000 \ C \ mol^{-1}$ $(i)$.Thus,the correct match is $a-iii, \ b-iv, \ c-ii, \ d-i$.

List-$I$	List-$II$
$a. \ Ag^{+}$	$i. \ 386000 \ C \ mol^{-1}$
$b. \ Mg^{2+}$	$ii. \ 289500 \ C \ mol^{-1}$
$c. \ Al^{3+}$	$iii. \ 96500 \ C \ mol^{-1}$
$d. \ Ti^{4+}$	$iv. \ 193000 \ C \ mol^{-1}$

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