In acidic solution, potassium manganate (K2MnO4) undergoes disproportionation: 3MnO4^2- + 4H^+ -> 2MnO4^- + MnO2 + 2H2O. The mole ratio of MnO4^- formed to MnO2 formed, and the oxidation states involved, are

2:1, with Mn(VI) going to Mn(VII) and Mn(IV)

1:2, with Mn(VI) going to Mn(VII) and Mn(IV)

3:1, with Mn(VI) going to Mn(VII) and Mn(II)

2:1, with Mn(VI) going to Mn(VII) and Mn(IV)

1:1, with Mn(VI) going to Mn(VII) and Mn(V)

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In acidic solution, potassium manganate ( $K X_{2} MnO X_{4}$ ) undergoes disproportionation: $3 MnO X_{4} X^{2 -} + 4 H X^{+} \to 2 MnO X_{4} X^{-} + MnO X_{2} + 2 H X_{2} O$ . The mole ratio of $MnO X_{4} X^{-}$ formed to $MnO X_{2}$ formed, and the oxidation states involved, are

A $1 : 2$ , with Mn(VI) going to Mn(VII) and Mn(IV)
B $3 : 1$ , with Mn(VI) going to Mn(VII) and Mn(II)
C $2 : 1$ , with Mn(VI) going to Mn(VII) and Mn(IV)
D $1 : 1$ , with Mn(VI) going to Mn(VII) and Mn(V)

Answer: C

In the balanced equation

2 MnO X_{4} X^{-}

and

1 MnO X_{2}

are formed. Each Mn(VI) that goes to Mn(VII) loses 1 electron; each that goes to Mn(IV) gains 2. Two losers per one gainer balance the electrons, so the ratio is

2 : 1

Practice d- and f-Block Elements More on Oxidation states and standard electrode potentials

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In acidic solution, potassium manganate (KX2​MnOX4​) undergoes disproportionation: 3MnOX4​X2−+4HX+→2MnOX4​X−+MnOX2​+2HX2​O. The mole ratio of MnOX4​X− formed to MnOX2​ formed, and the oxidation states involved, are

Answer: C

More questions on Oxidation states and standard electrode potentials

In acidic solution, potassium manganate ( $K X_{2} MnO X_{4}$ ) undergoes disproportionation: $3 MnO X_{4} X^{2 -} + 4 H X^{+} \to 2 MnO X_{4} X^{-} + MnO X_{2} + 2 H X_{2} O$ . The mole ratio of $MnO X_{4} X^{-}$ formed to $MnO X_{2}$ formed, and the oxidation states involved, are