Given the standard reduction potentials ${\rm{Z}}{{\rm{n}}^{2 + }}/{\rm{Zn}} = - 0.74V,{\rm{C}}{{\rm{l}}_2}/{\rm{C}}{{\rm{l}}^ - } = 1.36V$ ${{\rm{H}}^ + }/\frac{1}{2}{{\rm{H}}_2} = 0V{\kern 1pt} {\rm{and}}{\kern 1pt} {\rm{F}}{{\rm{e}}^{2 + }}/{\rm{F}}{{\rm{e}}^{3 + }} = 0.77V$ The order of increasing strength as reducing agent is

${\rm{C}}{{\rm{l}}^ - },{\rm{F}}{{\rm{e}}^{2 + }},{{\rm{H}}_2},{\rm{Zn}}$

${\rm{C}}{{\rm{l}}^ - },{\rm{Zn}},{{\rm{H}}_2},{\rm{F}}{{\rm{e}}^{2 + }}$

${{\rm{H}}_2},{\rm{Zn}},{\rm{F}}{{\rm{e}}^{2 + }},{\rm{C}}{{\rm{l}}^ - }$

${\rm{C}}{{\rm{l}}^ - },{\rm{F}}{{\rm{e}}^{2 + }},{\rm{Zn}},{{\rm{H}}_2}$

NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

1 Easy

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Understanding +4 | -1

Given the standard reduction potentials

\mathrm{Z}{\mathrm{n}^{2 + }}/\mathrm{Zn} = - 0.74V,\mathrm{C}{\mathrm{l}_2}/\mathrm{C}{\mathrm{l}^ - } = 1.36V

{\mathrm{H}^ + }/\frac{1}{2}{\mathrm{H}_2} = 0V{\kern 1pt} \mathrm{and}{\kern 1pt} \mathrm{F}{\mathrm{e}^{2 + }}/\mathrm{F}{\mathrm{e}^{3 + }} = 0.77V

The order of increasing strength as reducing agent is

View answer and explanation

Correct option: D

\mathrm{C}{\mathrm{l}^ - },\mathrm{F}{\mathrm{e}^{2 + }},{\mathrm{H}_2},\mathrm{Zn}

Option D (

\mathrm{C}{\mathrm{l}^ - },\mathrm{F}{\mathrm{e}^{2 + }},{\mathrm{H}_2},\mathrm{Zn}

) is correct because reducing strength increases as the standard reduction potential decreases.