In alkaline medium $H_{2} O_{2}$ reacts with $F e^{3 +}$ and $M n^{2 +}$ separately to give :

F e^{3 +}

and

M n^{4 +}

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F e^{2 +}

and

M n^{2 +}

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F e^{2 +}

and

M n^{4 +}

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F e^{3 +}

and

M n^{2 +}

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Solution

The correct option is C $F e^{2 +}$ and $M n^{4 +}$
In alkaline medium, $H_{2} O_{2}$ reduces $F e^{3 +}$ to $F e^{2 +}$ and oxidizes $M n^{2 +}$ to $M n^{4 +}$ .
$F e_{2} (S O_{4})_{3} + 2 K O H + H_{2} O_{2} \to 2 F e S O_{4} + K_{2} S O_{4} + 2 H_{2} O + O_{2}$
$M n S O_{4} + H_{2} O_{2} + 2 N a O H \to M n O_{2} + N a_{2} S O_{4} + 2 H_{2} O$

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Q. In alkaline medium,

H_{2} O_{2}

reacts with

F e^{3 +}

and

M n^{2 +}

separately to give:

E^{\circ}

for the half cells :

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O i s 1.23 V F e^{3 +} + e^{-} \to F e^{2 +} i s 0.77 V

Which of the following statements is true?

While $F e^{3 +}$ is stable, $M n^{3 +}$ is not stabe in acid solution because:
$(M n^{3 +} + e^{-} \to M n^{2 +}, E^{\circ} = 1.5 V; O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O, E^{\circ} = + 1.23 V; F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V)$

The balanced equation in basic medium is :

M n O_{4} + F e^{2 +} O H^{-} \to M n^{2 +} + F e^{3 +}

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In alkaline medium H2O2 reacts with Fe3+ and Mn2+ separately to give :

The correct option is C Fe2+ and Mn4+In alkaline medium, H2O2 reduces Fe3+ to Fe2+ and oxidizes Mn2+ to Mn4+.Fe2(SO4)3+2KOH+H2O2→2FeSO4+K2SO4+2H2O+O2MnSO4+H2O2+2NaOH→MnO2+Na2SO4+2H2O

In alkaline medium $H_{2} O_{2}$ reacts with $F e^{3 +}$ and $M n^{2 +}$ separately to give :