Question

The standard reduction potentials at 298 K for the following half-cell reactions are given:
\[Z{n^{2 + }}\left( {aq} \right) + 2{e^ - } \rightleftharpoons Zn\left( s \right);\; - 0.762\;{\text{V}}\]
\[C{r^{3 + }}\left( {aq} \right) + 3{e^ - } \rightleftharpoons Cr\left( s \right);\; - 0.74\;{\text{V}}\]
\[2{H^ + }\left( {aq} \right) + 2{e^ - } \rightleftharpoons {H_2}\left( s \right);\;\; + 0.00\;{\text{V}}\]
\[F{e^{3 + }}\left( {aq} \right) + {e^ - } \rightleftharpoons F{e^{2 + }}\left( {aq} \right);\;{\text{ }}\; + 0.77\;{\text{V}}\]

Which one of the following is the strongest reducing agent?
A.\[Zn\left( s \right)\]
B.\[Cr\left( s \right)\]
C.\[{H_2}\left( s \right)\]
D.\[F{e^{2 + }}\left( {aq} \right)\]

Answer 1

Hint: To answer this question, you should recall the concept of reduction potential. The electrode potential is oxidation potential and reduction potential termed as oxidation potential if oxidation takes place at the electrode. Reduction involves the gain of electrons, so the tendency of an electrode to gain electrons is called its reduction potential. The electrode having higher reduction potential always undergoes reduction during the reaction.


Complete Step by step solution:

Oxidation involves	Reduction involves
Losing electrons	Gaining electrons
Increase in oxidation number	A decrease in oxidation number
For a given compound losing hydrogen	For a given compound gaining hydrogen

When a piece of metal is immersed in a solution of its ions, a potential difference is created at the interface of the metal and the solution. The metal and ion represent the half cell and the reaction is half-reaction. The electrode having higher oxidation potential will be a stronger reducing agent than electrode having a lower oxidation potential. Hence electrodes having higher oxidation potential always undergo oxidation during the reaction. The electrode having the lowest reduction potential will be the strongest reducing agent than the electrode having lower reduction potential. Hence electrodes having higher reduction potential always undergo reduction during the reaction.
So, from the given options the reaction:
\[Z{n^{2 + }}\left( {aq} \right) + {2^{e - }} \rightleftharpoons Zn\left( s \right);{\text{ }} - 0.762\],
\[Zn\left( s \right)\] is the strongest reducing agent.

Therefore, the correct answer to this question is option A.

Note: We should know that at the electrodes, electrons are absorbed or released by the atoms and ions. There is gain or loss of electrons which are released and passed into the electrolyte. The uncharged atoms separate from the electrolyte upon exchanging electrons. This is called discharging and this concept helps determine the release of ions on different electrodes.
The increasing order of discharge of few cations is: \[{{\text{K}}^{\text{ + }}}{\text{, C}}{{\text{a}}^{{\text{2 + }}}}{\text{, N}}{{\text{a}}^{\text{ + }}}{\text{, M}}{{\text{g}}^{{\text{2 + }}}}{\text{, A}}{{\text{l}}^{{\text{3 + }}}}{\text{, Z}}{{\text{n}}^{{\text{2 + }}}}{\text{, F}}{{\text{e}}^{{\text{2 + }}}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, C}}{{\text{u}}^{{\text{2 + }}}}{\text{, A}}{{\text{g}}^{\text{ + }}}{\text{, A}}{{\text{u}}^{{\text{3 + }}}}\]Increasing Order of discharge of few anions is: \[{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{, N}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{, O}}{{\text{H}}^{\text{ - }}}{\text{, C}}{{\text{l}}^{\text{ - }}}{\text{, B}}{{\text{r}}^{\text{ - }}}{\text{, }}{{\text{I}}^{\text{ - }}}\]