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NEET UG CHEMISTRY MCQ: ELECTRO CHEMISTRY

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Standard reduction potentials of the half reactions are given below :

F2(g)+2e−→2F−(aq);E∘=+2.85V{\mathrm{F}_2}\left( \mathrm{g} \right) + 2{e^ - } \to 2{\mathrm{F}^ - }\left( {aq} \right);{E^ \circ } = + 2.85\mathrm{V}

Cl2(g)+2e−→2Cl−(aq);E∘=+1.36V\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) + 2{e^ - } \to 2\mathrm{C}{\mathrm{l}^ - }\left( {aq} \right);{E^ \circ } = + 1.36\mathrm{V}

Br2(l)+2e−→2Br−(aq);E∘=+1.06V\mathrm{B}{\mathrm{r}_2}\left( l \right) + 2{e^ - } \to 2\mathrm{B}{\mathrm{r}^ - }\left( {aq} \right);{E^ \circ } = + 1.06\mathrm{V}

I2(s)+2e−→2I−(aq);E∘=+0.53V{\mathrm{I}_2}\left( s \right) + 2{\mathrm{e}^ - } \to 2{\mathrm{I}^ - }\left( {aq} \right);{E^ \circ } = + 0.53V

The strongest oxidising and reducing agents respectively are :
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Correct option: A
F2andI−{\mathrm{F}_2}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\mathrm{I}^ - }
Option A (F2andI−{\mathrm{F}_2}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\mathrm{I}^ - }) is correct because F2 has the highest reduction potential (strongest oxidizing agent) and I- is the easiest to oxidize (strongest reducing agent).

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