$KMnO_4$ reacts with $H_2O_2$ in an acidic medium. The number of moles of oxygen produced per mole of $KMnO_4$ is

In the balanced reaction $x\,MnO_4^- + y\,H_2O \to 2Mn^{2+} + 5H_2O + 9O_2 + z\,e^-$,what are the values of $x$,$y$,and $z$?

When $KMnO_4$ acts as an oxidizing agent and ultimately forms $MnO_4^{2-}$,$MnO_2$,$Mn_2O_3$,and $Mn^{2+}$,what will be the number of electrons transferred in each case,respectively?

Given the redox reaction:
$X Na_2HAsO_3 + Y NaBrO_3 + Z HCl \longrightarrow NaBr + H_3AsO_4 + NaCl$
The values of $X$,$Y$,and $Z$ in the balanced equation are respectively:

In acidic medium,iodide $(I^{-})$ is oxidized to iodine $(I_2)$ by $KMnO_4$. What is the change in the oxidation number of manganese in this reaction?

In an experiment,$50 \ mL$ of $0.1 \ M$ solution of a metal salt reacted with $25 \ mL$ of $0.1 \ M$ solution of sodium sulphite. The half-equation for the oxidation of sulphite ion is $SO_3^{2-}{(aq)} + H_2O_{(l)} \to SO_4^{2-}{(aq)} + 2H^{+}{(aq)} + 2e^-$. If the oxidation number of the metal in the salt was $3$,what would be the new oxidation number of the metal?