What is obtained when solutions of $KI$ and $CuSO_4$ are mixed?

$CO_2$ and $SO_2$ gases can be distinguished by using:

Identify whether the following reactions act as oxidation or reduction processes:
$(i) \ S + 6HNO_3 \to H_2SO_4 + 6NO_2 + 2H_2O$
$(ii) \ 2K + Br_2 \to 2KBr$

Identify whether the following reactions act as oxidation or reduction processes:
$(i)$ $FeSO_4 + Mg \to MgSO_4 + Fe$
$(ii)$ $Cu + 4HNO_3 \to Cu(NO_3)_2 + 2NO_2 + 2H_2O$
$(iii)$ $H_2S + Cl_2 \to S + 2HCl$

Match the List-$I$ with List-$II$

List-$I$ $(\text{Redox Reaction})$	List-$II$ $(\text{Type of Redox Reaction})$
$A$. $CH_{4(g)} + 2O_{2(g)} \xrightarrow{\Delta} CO_{2(g)} + 2H_2O_{(l)}$	$I$. Disproportionation reaction
$B$. $2NaH_{(s)} \xrightarrow{\Delta} 2Na_{(s)} + H_{2(g)}$	$II$. Combination reaction
$C$. $V_2O_{5(s)} + 5Ca_{(s)} \xrightarrow{\Delta} 2V_{(s)} + 5CaO_{(s)}$	$III$. Decomposition reaction
$D$. $2H_2O_{2(aq)} \xrightarrow{\Delta} 2H_2O_{(l)} + O_{2(g)}$	$IV$. Displacement reaction

Choose the correct answer from the options given below:

Whenever a reaction between an oxidising agent and a reducing agent is carried out,a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.