Match the column :-Column I (Reduction proces | vedclass

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(B) The charge required for a reduction process is given by $Q = n 	imes F$,where $n$ is the number of moles of electrons transferred and $F = 96500

Vedclass · Accepted Answer

$(a)-(r), (b)-(s), (c)-(p), (d)-(q)$

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(B) The charge required for a reduction process is given by $Q = n 	imes F$,where $n$ is the number of moles of electrons transferred and $F = 96500 \ C/mol$.$(a)$ $MnO_4^- ightarrow Mn^{2+}$: Oxidation state of $Mn$ changes from $+7$ to $+2$. $n = 5$. Charge $= 5 	imes 96500 = 482500 \ C$. So,$(a)-(r)$.$(b)$ $Cr_2O_7^{2-} ightarrow 2Cr^{3+}$: Oxidation state of $Cr$ changes from $+6$ to $+3$. For $1$ mole of $Cr_2O_7^{2-}$,$n = 2 	imes 3 = 6$. Charge $= 6 	imes 96500 = 579000 \ C$. So,$(b)-(s)$.$(c)$ $Sn^{4+} ightarrow Sn^{2+}$: Oxidation state changes from $+4$ to $+2$. $n = 2$. Charge $= 2 	imes 96500 = 193000 \ C$. So,$(c)-(p)$.$(d)$ $Al^{3+} ightarrow Al$: Oxidation state changes from $+3$ to $0$. $n = 3$. Charge $= 3 	imes 96500 = 289500 \ C$. So,$(d)-(q)$.Correct mapping: $(a)-(r), (b)-(s), (c)-(p), (d)-(q)$.

Column $I$ (Reduction process)	Column $II$ (Charge required)
$(a)$ $1$ mol of $MnO_4^-$ to $Mn^{2+}$	$(p)$ $193000$ $C$
$(b)$ $1$ mole of $Cr_2O_7^{2-}$ to $Cr^{3+}$	$(q)$ $289500$ $C$
$(c)$ $1$ mole of $Sn^{4+}$ to $Sn^{2+}$	$(r)$ $482500$ $C$
$(d)$ $1$ mole of $Al^{3+}$ to $Al$	$(s)$ $579000$ $C$

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