In a weak alkaline medium,iodide is changed into which of the following compounds by $KMnO_4$?

In the given reaction,calculate the number of electrons lost by iron:
$3Fe + 4H_2O \to Fe_3O_4 + 4H_2$

For the redox reaction $MnO_4^- + C_2O_4^{2-} + H^+ \longrightarrow Mn^{2+} + CO_2 + H_2O$,the correct coefficients of the reactants for the balanced equation are $MnO_4^-$ - $C_2O_4^{2-}$ - $H^+$

Balance the following redox reactions by the ion-electron method:
$(a)$ $MnO_4^-(aq) + I^{-}(aq) \rightarrow MnO_2(s) + I_2(s)$ (in basic medium)
$(b)$ $MnO_4^-(aq) + SO_2(g) \rightarrow Mn^{2+}(aq) + HSO_4^-(aq)$ (in acidic solution)
$(c)$ $H_2O_2(aq) + Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + H_2O(l)$ (in acidic solution)
$(d)$ $Cr_2O_7^{2-} + SO_2(g) \rightarrow Cr^{3+}(aq) + SO_4^{2-}(aq)$ (in acidic solution)

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

In acidic medium,potassium dichromate acts as an oxidant according to the equation,$Cr_2O_7^{2-} + 14H^{+} + 6e^- \to 2Cr^{3+} + 7H_2O$. What is the equivalent weight of $K_2Cr_2O_7$? (Molecular weight $= M$)