What is the product of the oxidation of $I^-$ by $MnO_4^-$ in an alkaline medium?

Find the equivalent weight of $MnO_4^{-}$ for acid medium reactions. (Atomic weight of $Mn = 55$,Atomic weight of $O = 16$)

Balance the following redox reactions using the oxidation number and ion-electron method:
$(1) H_2S + Fe^{3+} \to Fe^{2+} + S + H^{+}$
$(2) Cu + NO_3^{-} \to Cu^{2+} + NO_2$
$(3) Sn + NO_3^{-} + H^{+} \to Sn^{2+} + NH_4^{+} + H_2O$
$(4) As + NO_3^{-} + H^{+} \to AsO_4^{3-} + NO_2 + H_2O$

After balancing the equation,$C_2O_4^{2-} + H^+ + MnO_4^- \longrightarrow CO_2 + Mn^{2+} + H_2O$,the coefficient of $CO_2$ is:

Number of moles of $H^{+}$ ions required by $1 \ mole$ of $MnO_4^{-}$ to oxidise oxalate ion to $CO_2$ is . . . . . .

Equivalent mass of the oxidizing agent in the reaction $SO_2 + 2H_2S \to 3S + 2H_2O$ is