\section{Important Points to Remember}

Redox titration : Potassium permanganate is often used in redox titrations because it is a powerful oxidizing agent and serves as its own indicator. In acidic solutions, the purple \(\mathrm{MnO}_{4}^{-}\)ion is reduced to the nearly colourless \(\mathrm{Mn}^{2+}\) ion.

- The oxidation number of \(\mathrm{N}\) in \(\mathrm{NO}_{2}\) is +4 . In \(\mathrm{NO}\), the oxidation number is +2 .

- \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a non-oxidizing acid when cold and dilute.

- The strongest oxidizing agent in a solution of a "nonoxidizing" acid is \(\mathrm{H}^{+}\).

- When we balance the equation, the ion-electron method will tell us how \(\mathrm{H}^{+}\)and \(\mathrm{H}_{2} \mathrm{O}\) are involved in the reaction.

- Electrons must be added to whichever side of the halfreaction is more positive (or less negative).

- The charge on the polyatomic ion equals the sum of the oxidation numbers of its atoms.

- Mostly, the medium in which a redox reaction is to be balanced is given in the problem but if the problem does not state the medium explicitly, then the medium is decided by looking at the reactants or products. If an acid or base is one of the reactants or products, then the medium is the same.

- \(\mathrm{KMnO}_{4}\) acts as an oxidant in every medium although with different strength which follows the order acidic medium \(>\) neutral medium \(>\) alkaline medium

- Act as both oxidising and reducing agents : \(\mathrm{SO}_{2}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{O}_{3}, \mathrm{NO}_{2}\), etc.

- Nature of oxides based on oxidation number

(i) Lowest oxidation state \(\rightarrow\) Basic (MnO)

(ii) Intermediate oxidation state \(\rightarrow\) Amphoteric \(\left(\mathrm{Mn}_{3} \mathrm{O}_{4}, \mathrm{MnO}_{2}\right)\)

(iii) Highest oxidation state \(\rightarrow\) Acidic \(\left(\mathrm{Mn}_{2} \mathrm{O}_{7}\right)\)

D Ion-electron method-Acidic solution for Balancing the equation

Step 1 : Divide the equation into two half-reactions.

Step 2 : Balance atoms other than \(\mathrm{H}\) and \(\mathrm{O}\).

Step 3 : Balance \(\mathrm{O}\) by adding \(\mathrm{H}_{2} \mathrm{O}\).

Step 4 : Balance \(\mathrm{H}\) by adding \(\mathrm{H}^{+}\).

Step 5 : Balance net charge by adding \(\mathrm{e}^{-}\).

Step 6 : Make \(\mathrm{e}^{-}\)gain equal to \(\mathrm{e}^{-}\)loss, then add halfreactions.

Step 7 : Cancel anything that's the same on both sides.

Additional steps in the ion-electron method for basic

solutions :

Step 8 : Add to both sides of the equation the same number of \(\mathrm{OH}^{-}\)as there are \(\mathrm{H}^{+}\).

Step 9 : Combine \(\mathrm{H}^{+}\)and \(\mathrm{OH}^{-}\)to form \(\mathrm{H}_{2} \mathrm{O}\).

Step 10 : Cancel any \(\mathrm{H}_{2} \mathrm{O}\) that you can.

D Fractional Oxidation State:

Fractional oxidation state is the average oxidation state of the element. eg. \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (tetrathionate ion), \(\mathrm{C}_{3} \mathrm{O}_{2}\) (carbon suboxide), and \(\mathrm{Br}_{3} \mathrm{O}_{8}\) (tribromooctaoxide)

D) Iodometric and Iodimetric titrations

\begin{tabular}{|c|l|c|}

\hline Estimation of & \multicolumn{1}{|c|}{ Reaction } & \(\begin{array}{c}\text { Reaction between oxidising } \\

\text { agent and reducing agent }\end{array}\) \\

\hline \(\mathrm{I}_{2}\) (Iodometry) & \(\begin{array}{l}\text { Titrating solution is } \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} .5 \mathrm{H}_{2} \mathrm{O}(\mathrm{Hypo}) \\

\mathrm{I}_{2}+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \rightarrow 2 \mathrm{NaI}_{2}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} \\

\text { or } \mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\end{array}\) & \(\begin{array}{l}\mathrm{I}_{2} \equiv 2 \mathrm{I}^{-} \equiv 2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \\

w\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\right)=\frac{M}{I}\end{array}\) \\

\hline \(\mathrm{CuSO}_{4}\) (Iodimetry) & \(\begin{array}{l}2 \mathrm{CuSO}_{4}+4 \mathrm{KI} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+2 \mathrm{~K}_{2} \mathrm{SO}_{4}+\mathrm{I}_{2} \\

\text { or } 2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2} \text { (white ppt.) }\end{array}\) & \(\begin{array}{l}2 \mathrm{CuSO}_{4} \equiv \mathrm{I}_{2} \equiv 2 \mathrm{I}^{-} \equiv 2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \\

E w \text { of } \mathrm{CuSO}_{4}=\frac{M}{I}\end{array}\) \\

\hline

\end{tabular}

(i) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(4 x-12=-2, x=2.5)\)

Oxidation number of \(\mathrm{S}=2.5\)

(ii) \(\mathrm{C}_{3} \mathrm{O}_{2}(3 x-4=0, x=4 / 3)\)

Oxidation number of \(\mathrm{C}=4 / 3\)

(iii) \(\mathrm{Br}_{3} \mathrm{O}_{8}(3 x-16=0, x=16 / 3)\)

Oxidation number of \(\mathrm{Br}=16 / 3\)