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Valency and Oxidation State

Equivalent we...

Question

Equivalent weight ${M n}^{3 +}$ in the following reaction is ( $M n = 55$ ) :
${M n}^{3 +} ⟶ {M n}^{2 +} + M n O_{2}$

27.5

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55

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110

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165

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Solution

The correct option is B $55$
${M n}^{3 +}$ disproportionates to ${M n}^{2 +}$ and $M n O_{2}$ (Mn(IV)).
Thus, equivalent mass of ${M n}^{3 +} = \frac{2 u n i t s o f {M n}^{3 +}}{2}$ $= \frac{M}{1} = 55$

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Similar questions

Q. Given the standard potential of the following at

25^{\circ} C

M n O_{2} \to M n^{3 +}

;

E^{⊖} = 0.95 V

M n^{3 +} \to M n^{2 +}

;

E^{⊖} = 1.51 V

The standard potential of

M n O_{2} \to M n^{2 +}

is:

The standard redox potential for reactions Mn²⁺+2e -----> Mn and Mn³⁺+1e -----> Mn²⁺ are 2.36 and 1.51V respectively . What is the redox potential for the reaction Mn³⁺+3e------> Mn ?

10

g of

M n O_{2}

on reaction with conc.

H C l

liberated

0.1

equivalent of

C l_{2}

(molecular weight of

M n = 55

g/mol). Hence, percentage purity of

M n O_{2}

is :

H_{2} O_{2}

reduces

M n O_{4}^{-}

ion to

Q. The

M n^{3 +}

ion is unstable in solution and undergoes disproportional reaction to give

M n^{2 +}, M n O_{2}

and

H^{+}

. Write a balanced ionic equation for the reaction.

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Equivalent weight Mn3+ in the following reaction is (Mn=55) :Mn3+⟶Mn2++MnO2

The correct option is B 55Mn3+ disproportionates to Mn2+ and MnO2 (Mn(IV)).Thus, equivalent mass of Mn3+=2unitsofMn3+2=M1=55

Equivalent weight ${M n}^{3 +}$ in the following reaction is ( $M n = 55$ ) :
${M n}^{3 +} ⟶ {M n}^{2 +} + M n O_{2}$

The correct option is B $55$
${M n}^{3 +}$ disproportionates to ${M n}^{2 +}$ and $M n O_{2}$ (Mn(IV)).
Thus, equivalent mass of ${M n}^{3 +} = \frac{2 u n i t s o f {M n}^{3 +}}{2}$ $= \frac{M}{1} = 55$