For the reaction M^{+n}_{(aq)} + ne^{-} right | vedclass

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(D) The reducing strength of a metal is directly proportional to its standard oxidation potential $(E^0_{oxi})$.The relationship between standard redu

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$M_2 > M_3 > M_1$

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(D) The reducing strength of a metal is directly proportional to its standard oxidation potential $(E^0_{oxi})$.The relationship between standard reduction potential $(E^0_{red})$ and standard oxidation potential $(E^0_{oxi})$ is: $E^0_{oxi} = -E^0_{red}$.Given values:For $M_1$: $E^0_{red} = -0.34 \ V \implies E^0_{oxi} = +0.34 \ V$For $M_2$: $E^0_{red} = -3.05 \ V \implies E^0_{oxi} = +3.05 \ V$For $M_3$: $E^0_{red} = -1.66 \ V \implies E^0_{oxi} = +1.66 \ V$Comparing the oxidation potentials: $3.05 \ V (M_2) > 1.66 \ V (M_3) > 0.34 \ V (M_1)$.Therefore,the order of reducing strength is $M_2 > M_3 > M_1$.

For the reaction $M^{+n}_{(aq)} + ne^{-} \rightarrow M_{(s)}$,if the standard reduction potentials of elements $M_1, M_2$,and $M_3$ are $-0.34 \ V$,$-3.05 \ V$,and $-1.66 \ V$ respectively,what is the order of their reducing strength?

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