Calculate the value of $n$ in the given reaction:
$MnO_4^{-} + 8H^{+} + n e^{-} \to Mn^{+2} + 4H_2O$

When chlorine gas is passed through an aqueous solution of $KBr$,the solution turns orange brown due to the formation of

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

In the reaction $Zn + 2H^{+} + 2Cl^{-} \to Zn^{2+} + 2Cl^{-} + H_2$,the spectator ion is

The following reaction occurs in acidic medium: $KMnO_4 + 8 H^{+} + 5 e^{-} \longrightarrow K^{+} + Mn^{2+} + 4 H_2O$. What is the equivalent weight of $KMnO_4$? (Molecular weight of $KMnO_4 = 158$)

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: