Question

Using the data given below find out the strongest reducing agent.

$E_{C{r}_2{O}_7^{2-}/C{r}^{3+}}^{\Theta }=1.33V$

$E_{C{l}_2/Cl}^{\Theta }=1.36V$

$E_{Mn{O}_4^{-}/M{n}^{2+}}^{\Theta }=1.51V$

$E_{C{r}^{3+}/Cr}^{\Theta }=-0.74V$

• A. $M{n}^{2+}$
• B. $Cr$
• C. $C{r}^{3+}$
• D. $C{l}^{-}$

Answer 1

The correct option is B $Cr$

Comparing standard reduction potential $(E^{\Theta }$

$E_{C{r}_2{O}_7^{2-}/C{r}^{3+}}^{\Theta }=1.33V$

$E_{C{l}_2/Cl}^{\Theta }=1.36V$

$E_{Mn{O}_4^{-}/M{n}^{2+}}^{\Theta }=1.51V$

\(E^{\Theta}_{Cr^{3+}/Cr}=-0.74V\

Higher the standard reduction potential, more is the tendency to get reduced (or lesser is the tendency to get oxidised).

Reducing power:

It is the ability of species to reduce others and oxidise itself.

Thus, the species with less reduction potential, will have more tendency to get oxidised and thus will have more reducing power or better will be the reducing agent.
Comparing the standard reduction potential $\left(E^{\Theta }\right)$

$E_{C{r}_2{O}_7^{2-}/C{r}^{3+}}^{\Theta }=1.33V$

$E_{C{l}_2/Cl}^{\Theta }=1.36V$

$E_{Mn{O}_4^{-}/M{n}^{2+}}^{\Theta }=1.51V$


$E_{C{r}^{3+}/Cr}^{\Theta }=-0.74V$

Higher the standard reduction potential, more is the tendency to get reduced (or lesser is the tendency to get oxidised).

Oxidising agent:

The species that oxidises others and gets reduced itself are called oxidising agents.

Thus, more is the reduction potential, more is the tendency to get reduced and thus more will be its oxidising power.

Comparing order of reducing power based on $E^{\Theta }$ values

Order of $E^{\Theta }$

$\left(Mn{O}_4^{-}/M{n}^{2+}\right)>\left(C{l}_2/C{l}^{-}\right)>\left(Cr{O}_7^{2-}/C{r}^{3+}\right)>\left(C{r}^{3+}/Cr\right)$

thus, order of reducing power:

$Mn2+<C{l}^{-}<C{r}^{3+}<Cr$

Therefore, $Cr$ is the strongest reducing agent.
Hence, correct answer is option (b).