Question

Using the data given, find out in which option the order of reducing power is correct.
$E_{C{r_2}O_7^{2 - }|C{r^{3 + }}}^\theta = 1.33V$
$E_{C{l_2}|C{l^ - }}^\theta = 1.36V$
$E_{MnO_4^ - \mid M{n^{2 + }}}^\theta = 1.51V$
$E_{C{r^{3 + }}\mid Cr}^\theta = - 0.74V$
A. $C{r^{3 + }} < C{l^ - } < M{n^{2 + }} < Cr$
B. $M{n^{2 + }} < C{l^ - } < C{r^{3 + }} < Cr$
C. $C{r^{3 + }} < C{l^ - } < C{r_2}O_7^{2 - } < Mn{O_4}^ -$
D. $M{n^{2 + }} < C{r^{3 + }} < C{l^ - } < Cr$

Answer 1

The standard electrode potential value is a measure of the tendency of a species to get reduced by accepting electrons. Reducing power is the exact opposite of this, and measures the tendency of a species to reduce other species, by getting oxidised itself. Thus, lower the electrode potential, higher the reducing power.

Complete step by step answer:
As we know, oxidation and reduction are inverse reactions of each other. While oxidation means removal of electrons, reduction is the gaining of electrons. Thus, when we say that a substance is a good reducing agent (has good reducing power), we actually mean that that substance can get oxidised easily. Therefore, a good reducing agent reduces another species by donating its electrons, and hence gets oxidised itself.
The standard electrode potential measures the tendency of a species to get reduced, that is, the tendency of the species to gain electrons. So, lower the electrode potential, lower is the tendency to get reduced, and consequently, higher is the tendency to get oxidised. Thus, species with lower electrode potential will get oxidised easily, can reduce other species and thus, have better reducing power (ability to reduce other species).
Looking at the data given, we see that the potential of $E_{C{r^{3 + }}\mid Cr}^\theta = - 0.74V$ is the least ($- 0.74V$). Thus, this will have the highest reducing power. The next smallest is for $E_{C{r_2}O_7^{2 - }|C{r^{3 + }}}^\theta = 1.33V$, followed by $E_{C{l_2}|C{l^ - }}^\theta = 1.36V$ and lastly, $E_{MnO_4^ - \mid M{n^{2 + }}}^\theta = 1.51V$.
Therefore, the increasing order (smallest to largest) of reducing power is:
$M{n^{2 + }} < C{l^ - } < C{r^{3 + }} < Cr$
Therefore, the correct option is B.

Note: The standard electrode potential commonly used is also known as the standard reduction potential, as it measures the ability of a species to get reduced. In this series, hydrogen is taken as the reference and is assigned a reduction potential of zero. Note that the oxidising power is the inverse of reducing power, and therefore, increases as the reduction potential increases. There is also a standard oxidation potential series, which measures the ability of a species to get oxidised.