In balancing the half-reaction $S_2O_3^{2-} \to S_{(s)}$,the number of electrons that must be added is:

The product of oxidation of $I^{-}$ with $MnO_4^{-}$ in alkaline medium is

Balance the following redox reactions by the ion-electron method:
$(a)$ $MnO_{4(aq)}^- + I_{(aq)}^- \to MnO_{2(s)} + I_{2(s)}$ (in basic solution)
$(b)$ $MnO_{4(aq)}^- + SO_{2(g)} \to Mn_{(aq)}^{2+} + HSO_{4(aq)}^-$ (in acidic solution)
$(c)$ $H_2O_{2(aq)} + Fe_{(aq)}^{2+} \to Fe_{(aq)}^{3+} + H_2O_{(l)}$ (in acidic solution)
$(d)$ $Cr_2O_7^{2-} + SO_{2(g)} \to Cr_{(aq)}^{3+} + SO_{4(aq)}^{2-}$ (in acidic solution)

$MnO_4^{-}$ oxidises $(i)$ oxalate ion in acidic medium at $333 \ K$ and $(ii)$ $HCl$. For balanced chemical equations,the ratios $[MnO_4^{-} : C_2O_4^{2-}]$ in $(i)$ and $[MnO_4^{-} : HCl]$ in $(ii)$ respectively are

$NaClO$ solution reacts with $H_2SO_3$ as,$NaClO + H_2SO_3 \to NaCl + H_2SO_4$. $A$ solution of $NaClO$ used in the above reaction contained $15 \ g$ of $NaClO$ per litre. The normality of the solution would be

In alkaline medium,the reduction of permanganate anion involves a gain of $........$ electrons.