Calculate the value of $n$ in the given reaction:
$MnO_4^{-} + 8H^{+} + n e^{-} \to Mn^{+2} + 4H_2O$

In an experiment,$50 \ mL$ of $0.1 \ M$ solution of a metal salt reacted with $25 \ mL$ of $0.1 \ M$ solution of sodium sulphite. The half-equation for the oxidation of sulphite ion is $SO_3^{2-}{(aq)} + H_2O_{(l)} \to SO_4^{2-}{(aq)} + 2H^{+}{(aq)} + 2e^-$. If the oxidation number of the metal in the salt was $3$,what would be the new oxidation number of the metal?

Permanganate $(VII)$ ion,$MnO_4^-$,in basic solution oxidises iodide ion,$I^-$,to produce molecular iodine $(I_2)$ and manganese $(IV)$ oxide $(MnO_2)$. Write a balanced ionic equation to represent this redox reaction.

Which of the following equations is a balanced one?

When $SO_{2}$ is passed through acidified $K_{2}Cr_{2}O_{7}$,the process that takes place is

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