For a redox reaction,how many moles of electrons will one mole of $MnO_4^-$ accept in an acidic medium?

In the given half-reaction $Cr_2O_7^{2-} + ZH^{+} + e^- \to Cr^{3+} + H_2O$,find the value of $Z$.

Write the balanced chemical equation for the following reaction using the oxidation number method:
Reaction of liquid hydrazine $(N_2H_4)$ with chlorate ion $(ClO_3^-)$ in basic medium produces nitric oxide gas $(NO)$ and chloride ion $(Cl^-)$.

For the redox reaction $ xMnO_{4}^{-} + yH_{2}C_{2}O_{4} + zH^{+} \rightarrow mMn^{2+} + nCO_{2} + pH_{2}O $,the values of $ x, y, m $ and $ n $ are:

Permanganate $(VII)$ ion,$MnO_4^-$,in basic solution oxidises iodide ion,$I^-$,to produce molecular iodine $(I_2)$ and manganese $(IV)$ oxide $(MnO_2)$. Write a balanced ionic equation to represent this redox reaction.

For the reaction:
$I^{-} + ClO_{3}^{-} + H_{2}SO_{4} \longrightarrow Cl^{-} + HSO_{4}^{-} + I_{2}$
The correct statement$(s)$ in the balanced equation is/are:
$A$. Stoichiometric coefficient of $HSO_{4}^{-}$ is $6$.
$B$. Iodide is oxidized.
$C$. Sulphur is reduced.
$D$. $H_{2}O$ is one of the products.