`AO_(2)` disproportionates into `AO_(4)^(-1)` and `A^(n+)` ion. If the mole ratio of `AO_(2)` undergone oxidation and reduction is `2:3`, the value of `n` is

\(AO_{2}\) disproportionates in to \(AO_{4}^{-1}\) and \(A^{ n+}\) ion. If the mole ratio of \(AO_{2}\) undergone oxidation and reduction is 2:3, the value of "n" is

One mole of MnO_(4)^(2-) disproportionates to yield :

In the disproportionation reaction ( unbalanced ) , Br_(2) + OH^(-) rarr Br^(-) + BrO_(3)^(-) + H_(2)O The ratio of Br_(2) molecules undergoing oxidation and reduction is "P/Q" then "P+Q" is

3MnO_2 + 4Al to 3Mn + 2Al_2O_3 + Heat Identify the substances undergone oxidation and reduction in this reaction

3MnO_(2) + 4Al rarr 3 Mn + 2Al_(2)O_(3)+ heat. Identify the substances undergone oxidation and reduction reactions.

In the reaction, A^(n+)+MnO_(4)^(-) rarr A^(5) +Mn^(2+) if 0 0.05 mole of A^(n+) is oxidized by 0.02 mole of MnO_(4)^(-) , the value of 'n' is