What is the equivalent weight of $IO_4^-$ when it is converted to $I_2$ in an acidic medium?

The number of electrons involved in the reduction of permanganate to manganese dioxide in acidic medium is $.......$.

In the redox reaction
$x \ MnO + y \ PbO_2 + z \ HNO_3 \to a \ HMnO_4 + b \ Pb(NO_3)_2 + c \ H_2O$
Find the values of $x, y, z$.

After balancing the reaction $Cu + HNO_3 \rightarrow Cu(NO_3)_2 + NO_2 + H_2O$ in an acidic medium,the number of nitrogen atoms,the number of water molecules,and the net charge on the product side are respectively:

$2 MnO_{4}^{-} + b C_{2}O_{4}^{2-} + c H^{+} \rightarrow x Mn^{2+} + y CO_{2} + z H_{2}O$
If the above equation is balanced with integer coefficients,the value of $c$ is ...........
(Round off to the Nearest Integer).

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 $KMnO_4 = 158$)