Balance the following redox reactions using the oxidation number and ion-electron method:
$(1) \ Zn + NO_3^- + H^+ \to Zn^{2+} + N_2O + H_2O$
$(2) \ I^- + O_2 + H_2O \to I_2 + OH^-$
$(3) \ MnO_4^- + C_2H_5OH \to Mn^{2+} + CH_3COOH$

Which of the following are redox changes?
$(i)$ Hydrolysis of $XeF_2$
$(ii)$ Thermal decomposition of $Ag_2CO_3$
$(iii)$ Reaction of $Al$ with $NaOH$
$(iv)$ Hydrolysis of $PCl_5$
$(v)$ Reaction of $NaCl$ with conc. $H_2SO_4$
Correct options are:

$50 \ mL$ of $0.1 \ M$ solution of a salt reacted with $25 \ mL$ of $0.1 \ M$ solution of sodium sulphite. The half-reaction for the oxidation of sulphite ion is:
$SO_{3}^{2-} (aq) + H_{2}O (l) \to SO_{4}^{2-} (aq) + 2H^{+} (aq) + 2e^{-}$
If the oxidation number of the metal in the salt was $+3$,what would be the new oxidation number of the metal?

How many moles of $H_2SO_3$ are required to reduce one mole of $KMnO_4$ in an acidic medium?

$40 \text{ mL}$ of $x \text{ M } KMnO_4$ solution is required to react completely with $200 \text{ mL}$ of $0.02 \text{ M}$ oxalic acid solution in acidic medium. The value of $x$ is:

Aqueous sulphite reacts with dilute sulphuric acid to form $X_{(g)}$. The liberated $X_{(g)}$ is passed into acidified $KMnO_4$ solution. What is the oxidation state of $Mn$ in the product formed?