Question

The different oxidation state which show spontaneous disproportionation in acidic medium are:

• A. $M{n}^{6+}$ and $M{n}^{3+}$
• B. $M{n}^{6+}$ and $M{n}^{2+}$
• C. $M{n}^{7+}$ and $M{n}^{3+}$
• D. $M{n}^{4+}$ and $M{n}^{2+}$

Answer 1

The correct option is A $M{n}^{6+}$ and $M{n}^{3+}$

A particular oxidation state will show disproportionation reaction if cell formed by this will have positive emf value.
So $M{n}^{+6}$ will show as its emf is positive.
$3M{n}^{6+}\to 2M{n}^{7+}+M{n}^{4+}$;

$E^o=E_{O{P}_{M{n}^{6+}/M{n}^{7+}}}^o+E_{R{P}_{M{n}^{6+}/M{n}^{4+}}}^o$

$=-0.56+2.26$

$=1.70V$
Also, $Mn+3$ will show such reaction as its emf is also positive as
$2M{n}^{3+}\to M{n}^{4+}+M{n}^{2+}$;

$E^o=E_{O{P}_{M{n}^{3+}/M{n}^{4+}}}^o+E_{R{P}_{M{n}^{3+}/M{n}^{2+}}}^o$

$=-0.95+1.51$

$=+0.56V$