Name: Redox Reactions as a Basis for Titration
Uploaded: 2019-07-19
Duration: 7 min 17 s
Description: This video will help us learn about how redox indicators like potassium permanganate or diphenylamine are used to indicate the completion of a redox reaction...

Question

Starting with the correctly balanced half-reactions, write the overall net ionic reaction in the following change:

Nitrous acid $(H N O_{2})$ Oxidises $I^{-}$ to $I_{2}$ (in acid solution). Also, find out the change in the oxidation number of the underlined atom.

Accepted Answer

For balancing the given reaction,

Step 1. First, write the skeletal equation.

$I^{-} + {HNO}_{2} + H^{+} \to NO + H_{2} O + I_{2}$

Step.2 Write the oxidation number of all the elements above their respective symbols.

$\overset{- 1}{I^{-}} + \overset{+ 1}{H} \overset{+ 3}{N} {\overset{- 2}{O}}_{2} + \overset{+ 1}{H^{+}} \to \overset{+ 2}{N} \overset{- 2}{O} + {\overset{+ 1}{H}}_{2} \overset{- 2}{O} \overset{}{+ \overset{0}{I_{2}}}$

Step.3 Find out the species which have been oxidised and reduced and split the given skeletal equation into two half-reactions.

Since the oxidation number of $N$ decreases from $+ 3$ in ${HNO}_{2}$ to $+ 2$ in $NO$ and that of $(H N O_{2})$ is increases from $- 1$ in $I^{-}$ to $0$ in $I_{2}$ . Thus, the above equation can be divided into two half-reactions.

Oxidation half equation: $I^{-} \to I_{2} \dots \dots \dots \dots \dots \dots \dots (i)$

Reduction half equation : ${NO}_{2}^{-} \to NO \dots \dots \dots \dots \dots \dots (ii)$

Step.4 Balance oxidation and reduction half-reactions both.

$(a)$ Balance all the atoms other than $O$ and $H$ .

$2 I^{-} \to I_{2} \dots \dots \dots \dots \dots \dots (iii)$

$N$ atom is already balanced.

$(b)$ Balance $O$ atoms by adding $H_{2} O$ .

${NO}_{2}^{-} \to NO + H_{2} O \dots \dots \dots \dots \dots (iv)$

As eq. $(iii)$ does not contain $O$ and $H$ , so no need to balance oxygen and hydrogen.

$(c)$ Balance $H$ atoms by adding $H^{+}$ .

${NO}_{2}^{-} + 2 H^{+} \to NO + H_{2} O \dots \dots \dots \dots (v)$

$(d)$ Now, balance the charge of both sides.

$2 I^{-} \to I_{2} + 2 e^{-} \dots \dots \dots \dots \dots \dots \dots (vi)$

${NO}_{2}^{-} + 2 H^{+} + e^{-} \to NO + 2 H_{2} O \dots \dots \dots \dots (vii)$

Step.5 To balance the electrons lost in the equation $(vi)$ and gained in equation $(vii)$ , multiply eq. $(vii)$ by $2$ and add both equations.

$2 I^{-} \to I_{2} + 2 e^{-}$

$2 {NO}_{2}^{-} + 4 H^{+} + 2 e^{-} \to 2 NO + 2 H_{2} O$ ____________________________________

Thus, the overall balanced reaction is
$2 I^{-} + 4 H^{+} + 2 {NO}_{2}^{-} \to 2 NO + 2 H_{2} O + I_{2}$

And the oxidation number of $N$ from $+ 3$ in ${HNO}_{2}$ to $+ 2$ in $NO$ .

Starting with the correctly balanced half-reactions, write the overall net ionic reaction in the following change:

Nitrous acid $(H N O_{2})$ Oxidises $I^{-}$ to $I_{2}$ (in acid solution). Also, find out the change in the oxidation number of the underlined atom.

Important Questions on Redox Reactions

Learn and Prepare for any exam you want !

Starting with the correctly balanced half-reactions, write the overall net ionic reaction in the following change: Nitrous acid HNO2 Oxidises I- to I2 (in acid solution). Also, find out the change in the oxidation number of the underlined atom.

Important Questions on Redox Reactions

Learn and Prepare for any exam you want !

Starting with the correctly balanced half-reactions, write the overall net ionic reaction in the following change:

Nitrous acid $(H N O_{2})$ Oxidises $I^{-}$ to $I_{2}$ (in acid solution). Also, find out the change in the oxidation number of the underlined atom.