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MnO(4)^(2-) + H^(+) = MnO(4)^(-) + MnO(2...

`MnO_(4)^(2-) + H^(+) = MnO_(4)^(-) + MnO_(2) + H_(2) O`

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To balance the redox reaction \( \text{MnO}_4^{2-} + \text{H}^+ \rightarrow \text{MnO}_4^{-} + \text{MnO}_2 + \text{H}_2\text{O} \), we will follow these steps: ### Step 1: Determine the oxidation states - In \( \text{MnO}_4^{2-} \), the oxidation state of Mn can be calculated as follows: \[ x + 4(-2) = -2 \implies x - 8 = -2 \implies x = +6 \] - In \( \text{MnO}_4^{-} \), the oxidation state of Mn is: ...
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Which of the following reactions are disproportionation reactions ? (i) Cu^(+) to Cu^(2+) + Cu (ii) 3MnO_4^(2-) + 4H^(+ ) to 2MnO_4^(-) + MnO_2 + 2H_2O (iii) 2KMnO_4 to K_2 MnO_4 + MnO_2 + O_2 (iv) 2MnO_4^(-) + 3Mn^(2+) + 2H_2O to 5MnO_2 + 4H^+

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MnO_(4)^(-) is good oxidising agent in different medium changing to - rarr MnO_(4)^(2-) rarr MnO_(4)^(2-) rarr MnO_(2) rarr Mn_(2)O_(3) Changes in oxidation number respectively are -

KMnO_(4) can be prepared from K_(2)MnO_(4) as per the reaction 3MnO_(4)^(2-) + 2H_(2)O hArr 2MnO_(4)^(2-) + MnO_2 + 4OH^- The reaction can go to completion by removing OH^(-) ions by adding.