When $H_2S$ is passed through acidified $KMnO_4,$ we get

The reaction,$3 ClO^{-} \rightarrow ClO_3^{-} + 2 Cl^{-}$ occurs in two steps:
$i.$ $2 ClO^{-} \rightarrow ClO_2^{-} + Cl^{-}$
$ii.$ $ClO_2^{-} + ClO^{-} \rightarrow ClO_3^{-} + Cl^{-}$
The reaction intermediate is:

What will be the coefficients $x, y, z$ in the balanced form of the following redox equation: $xIO_3^- + yHSO_3^- \to zI^{-} + pH^{+} + qSO_4^{2-}$?

What is the value of $x$ and $y$ in order to balance the following redox reaction?
$x CuO + y NH_3 \longrightarrow x Cu + N_2 + x H_2 O$

$KMnO_4 + HCl \rightarrow MnCl_2 + Cl_2 + KCl + H_2O$. In the above reaction,how many moles of $H_2O$ would be formed for each mole of $Cl_2$ liberated?

The product of $I^-$ with $MnO_4^-$ in acidic medium is