Only $2 \ mL$ of $KMnO_4$ solution of unknown molarity is required to reach the end point of a titration of $20 \ mL$ of oxalic acid $(2 \ M)$ in acidic medium. The molarity of $KMnO_4$ solution should be . . . . . . . . . $M$.

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
$(a)$ $P_{4(s)} + OH_{(aq)}^{-} \to PH_{3(g)} + HPO_{2(aq)}^{-}$
$(b)$ $N_{2}H_{4(l)} + ClO_{3(aq)}^{-} \to NO_{(g)} + Cl_{(g)}^{-}$
$(c)$ $Cl_{2}O_{7(g)} + H_{2}O_{2(aq)} \to ClO_{2(aq)}^{-} + O_{2(g)} + H^{+}$

$KMnO_4$ oxidises oxalic acid to $CO_2$ in acidic medium. The equivalent weight of $KMnO_4$ is $(Mn = 55, K = 39, O = 16)$:

Consider the following statements:
In the chemical reaction:
$MnO_2 + 4HCl \to MnCl_2 + 2H_2O + Cl_2$
$(1)$ Manganese ion is oxidised
$(2)$ Manganese ion is reduced
$(3)$ Chloride ion is oxidised
$(4)$ Chloride ion is reduced.
Which of these statements are correct?

In alkaline condition $KMnO_4$ reacts as follows:
$2KMnO_4 + 2KOH \to 2K_2MnO_4 + H_2O + O$
Therefore,its equivalent weight will be:

Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.