Question

Which is the Strongest Reducing Agent?

• A. $\mathrm{Zn}\left(\mathrm{s}\right)$
• B. $\mathrm{Cr}\left(\mathrm{s}\right)$
• C. ${\mathrm{H}}_2\left(\mathrm{g}\right)$
• D. ${\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)$

Answer 1

The correct option is A

$\mathrm{Zn}\left(\mathrm{s}\right)$

Explanation for the correct options:

Metals are generally good reducing agents. Reduction is gain of electrons. When a metal loses its electrons it oxidizes itself and reduces others.
The more easily a metal loses its electrons, the better it is as a reducing agent.

"The more effective the oxidizing agent, the higher the electrode potential."
Because electrode potential decreases in the following order:

• ${\mathrm{H}}^{+}/{\mathrm{H}}_2>{\mathrm{Fe}}^{2+}/\mathrm{Fe}>{\mathrm{Cr}}^{3+}/\mathrm{Cr}>{\mathrm{Zn}}^{2+}/\mathrm{Zn}$

Zn(S) has a higher reducing power as a result. As a result, the correct response is Option (A).

Explanation for the incorrect options:

(B) It has lower electrode potential then $\underset{\mathrm{Reduction}\mathrm{potential}}{\underbrace{\mathrm{Cr}<\mathrm{Zn}}}$.

(c) It has lower electrode potential then $\underset{\mathrm{Reduction}\mathrm{potential}}{\underbrace{{\mathrm{H}}_2<\mathrm{Zn}}}$

(D) It has lower electrode potential then $\underset{\mathrm{Reduction}\mathrm{potential}}{\underbrace{{\mathrm{Fe}}^{+2}<\mathrm{Zn}}}$..

Hence, the correct option would be.option (A)