In the half-reaction $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$,the equivalent weight of $Cr_2O_7^{2-}$ is obtained by dividing its molecular weight by:

In the balanced reaction $x\,MnO_4^- + y\,H_2O \to 2Mn^{2+} + 5H_2O + 9O_2 + z\,e^-$,what are the values of $x$,$y$,and $z$?

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)

In the reaction $I_2 + 2S_2O_3^{2-} \to 2I^{-} + S_4O_6^{2-}$,the equivalent weight of iodine will be equal to

One of the products formed due to the reaction between $KMnO_4$ and $HCl$ is

$Na_2S_2O_3 + I_2 \to$ Product is