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 two illustrations.

Consider the elements :
$\mathrm{C}\mathrm{s}, \mathrm{N}\mathrm{e}, \mathrm{I}$ and $\mathrm{F}$

Identify the element that exhibits only negative oxidation state, identify the element that exhibits only positive oxidation state, identify the element that exhibits both positive and negative oxidation states and identify the element which exhibits neither the negative nor does the positive oxidation state.

Justify that the following reaction are redox reaction:

${\mathrm{CuO}}_{\left(\mathrm{s}\right)} + {\mathrm{H}}_{2\left(\mathrm{g}\right)} \to {\mathrm{Cu}}_{\left(\mathrm{s}\right)} + {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$

Justify that the following reactions are redox reactions;

${\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+3\mathrm{CO}\left(\mathrm{g}\right)\to 2\mathrm{Fe}\left(\mathrm{s}\right)+3{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Justify that the following reactions are redox reactions:

${\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)+3\mathrm{CO}\left(\mathrm{g}\right)\to 2\mathrm{Fe}\left(\mathrm{s}\right)+3{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Balance the following reaction by the ion-electron method:

${{\mathrm{MnO}}_4}^{-}\left(\mathrm{aq}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right) \to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+{{\mathrm{HSO}}_4}^{-}\left(\mathrm{aq}\right)$ (in acidic solution)