Question

Using the data given in below find out in which option the order of reducing power is correct.
$E_{C{r}_2{O}_7^{2-}/C{r}^{3+}}^{\circ }=1.33V$ ; $E_{C{l}_2/C{l}^{-}}^{\circ }=1.36V$
$E_{Mn{O}_4^{-}/M{n}^{2+}}^{\circ }=1.51V$ ; $E_{C{r}^{3+}/Cr}^{\circ }=-0.74V$

• A. $C{r}^{3+}<C{l}^{-}<M{n}^{2+}<Cr$
• B. $M{n}^{3+}<C{l}^{-}<C{r}^{3+}<Cr$
• C. $C{r}^{3+}<C{l}^{-}<C{r}_2{O}_7^{2-}<Mn{O}_4^{-}$
• D. $M{n}^{2+}<C{l}^{-}<C{r}^{3+}<Cr$

Answer 1

Answer: (D)

Electrochemical series is a series of chemical elements arranged in order of their standard electrode potentials.

Down the electrochemical series the reduction potential decreases and more the reduction potential lower the reducing power and less the reduction potential, higher the reducing power.
So the option $B,M{n}^{2+}<C{l}^{-}<C{r}^{3+}<Cr$ is the decreasing order of reduction potential and hence is the correct option.