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In acidic medium, potassium manganate un...

In acidic medium, potassium manganate undergoes disproportionation according to equation.\(3MnO_{4}^{2-}+4H^{+}\rightarrow 2MnO_{4}^{-}+MnO_{2}+2H_{2}O\)
Equivalent weight of \(K_{2}MnO_{4}\) will be (Mol.wt.of \(K_{2}MnO_{4}\)=M)

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In acidic medium,potassium manganate undergoes disproportionation according to equation. 3MnO_(4)^(2-)+4H^(+)longrightarrow2MnO_(4)^(-)+MnO_(2)+2H_(2)O Equivalent weight of K_(2)MnO_(4) will be (Mol.wt.of K_(2)MnO_(4) =M)

In acidic medium MnO_(4)^(2-)

Potassium manganate (K_2MnO_4) is formed when

The equation 3MnO_4^(2-)+4H^(+)to2MnO_4^(-)+MnO_2+2H_2O represents

Write the disproportionate reaction of MnO_(4)^(2-) ?

The equivalent weight of MnO_(4)^(-) ion in acidic medium is