For a redox reaction,how many moles of electrons will one mole of $MnO_4^-$ accept in an acidic medium?

$aIO_3^- + bI^- + 6H^+ \to cI_2 + 3H_2O$
In the above reaction,the coefficients $a$,$b$,and $c$ are respectively:

The value of $x$ in the partial redox equation $MnO_4^- + 8H^{+} + xe^- \rightleftharpoons Mn^{2+} + 4H_2O$ is

Balance the following ionic equations:
$1. Cr_{2}O_{7}^{2-} + H^{+} + I^{-} \to Cr^{3+} + I_{2} + H_{2}O$
$2. Cr_{2}O_{7}^{2-} + Fe^{2+} + H^{+} \to Cr^{3+} + Fe^{3+} + H_{2}O$
$3. MnO_{4}^{-} + SO_{3}^{2-} + H^{+} \to Mn^{2+} + SO_{4}^{2-} + H_{2}O$
$4. MnO_{4}^{-} + H^{+} + Br^{-} \to Mn^{2+} + Br_{2} + H_{2}O$

When neutral or faintly alkaline $KMnO_4$ is treated with potassium iodide,iodide ion is converted into '$X$'. '$X$' is

In the reaction $A^{-n_2} + x e^- \to A^{-n_1}$,the value of $x$ is: