Equivalent mass of Ba ( MnO 4 ) 2 in acidic medium is : (where M stands for molar mass)

Question

Equivalent mass of Ba ( MnO 4 ​ ) 2 ​ in acidic medium is : (where M stands for molar mass)

TestHub · Accepted Answer

Explain the question : Calculate equivalent mass of $Ba (MnO_{4})_{2}$ .

Concept: This question is based on Equivalent mass / change in oxidation state

Solution: The compound is barium permanganate $Ba (MnO_{4})_{2}$ . In this salt, Barium (Ba) has an oxidation state of +2 , and the permanganate ion $(MnO_{4}^{-})$ is present. In $MnO_{4}^{-}$ , the oxidation state of manganese $(Mn)$ is calculated as : $x + 4 (- 2) = - 1 \Rightarrow x - 8 = - 1 \Rightarrow x = + 7$ . In an acidic medium, the permanganate ion $(MnO_{4}^{-})$ acts as a strong oxidizing agent and is reduced to the manganese (II) ion $(Mn^{2 +})$ . The halfreaction is:

$MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \to Mn^{2 +} + 4 H_{2} O$

The change in the oxidation state of Mn is from +7 to +2 , which corresponds to a change of 5 units per $MnO_{4}^{-}$ ion.

The $Ba (MnO_{4})_{2}$ molecule contains two $MnO_{4}^{-}$ ions. Therefore, the total change in oxidation state (or the n -factor) for the molecule is $2 \times 5 = 10$

The equivalent mass (E) is given by the formula:

$E = \frac{Molar mass (M)}{n - factor} = \frac{M}{10}$

Final answer : C

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