Question

Use the data given, find out the most stable ion in its reduced form.
$E_{C{r}_2{O}_7^{2-}|C{r}^{3+}}^{\theta }=1.33V$
$E_{C{r}_2|C{l}^{-}}^{\theta }=1.36V$
$E_{Mn{O}_4^{-}|{Mn}^{2+}}^{\theta }=1.51V$
$E_{C{r}^{3+}|Cr}^{\theta }=-0.74V$

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

Answer 1

Answer: (D)

$M{n}^{2+}$

When we have a look at Reduction potentials of following,

$E_{C{l}^{-}/Cl}^0=1.36V$

$E_{C{r}^{3+}/Cr}^0=-0.74V$

$E_{Cr}^0=0.74V$

$E_{Mn{O}_4^{-}/M{n}^{2+}}^0=1.51V$

As the element having higher reduction potential is most stable in its reduced form we have $E_{Mn{O}_4^{-}/M{n}^{2+}}^0$ having highest reduction potential and is the right option.