Write the O. N. of all atoms above their respective symbols.

O. N. decreases by, 3 per Cr-atom

Divide the given equation into two half reactions

Reduction half reaction: $\mathrm{Cr}_2 \mathrm{O}_7 \rightarrow \mathrm{Cr}^{3+}$

Oxidation half reaction: $\mathrm{I}^{-} \rightarrow \mathrm{I}_2$

To balance reduction half reaction.

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$$

To balance oxidation half reaction

$$2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 \mathrm{e}^{-}$$

To balance the reaction by electrons gained and lost

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$$

$6 \mathrm{I}^{-} \longrightarrow 3 \mathrm{I}_2+6 \mathrm{e}^{-}$

$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}$

This gives the final balanced ionic equations. (b) Write the skeletal equation of the given reaction

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Fe}^{3+}(a q)$$

Write the O. N. of all the elements above their respective symbols.

Divide the given equation into two half reactions Oxidation half reaction:

$\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)$ reduction half reaction : $\mathrm{Cr}_2 \mathrm{O}_7^{2-}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(a q)$

To balance oxidation half reaction

$$\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-}$$

To balance reduction half reaction

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}(a q)+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}(a q)$$

Balance charge by adding $\mathrm{H}^{+}$ ions.

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}(\mathrm{aq})+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}(a q)$$

Balance O atoms by adding $\mathrm{H}_2 \mathrm{O}$ molecules

$$\mathrm{Cr}_2 \mathrm{O}_7^{2-}(a q)+14 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_2 \mathrm{O}(l)$$

To balance the reaction

$$\begin{gathered} 6 \mathrm{Fe}^{2+}(a q) \longrightarrow 6 \mathrm{Fe}^{3+}(a q)+6 \mathrm{e}^{-} \\ \mathrm{Cr}_2 \mathrm{O}_7^{2-}(a q)+14 \mathrm{H}^{+}(a q)+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_2 \mathrm{O}(l) \end{gathered}$$

(c) Write the O.N. of all atoms above their respective symbols.

Divide the skeleton equation into two half-reactions.

Reduction half reaction: $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$

Oxidation half reaction : $\mathrm{SO}_3^{2-} \longrightarrow \mathrm{SO}_4^{2-}$

To balance reduction half reaction

$$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$$

To balance oxidation half reaction

$$\mathrm{SO}_3^{2-} \longrightarrow \mathrm{SO}_4^{2-}+2 e^{-}$$

Balance charge by adding $\mathrm{H}^{+}$ions.

$$\mathrm{SO}_3^{2-} \longrightarrow \mathrm{SO}_4^{2-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$$

Balance O -atoms by adding $\mathrm{H}_2 \mathrm{O}$ molecules

$$\mathrm{SO}_3^{2-}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{SO}_4^{2-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$$

To balance the reaction

$$\begin{aligned} & 2 \mathrm{MnO}_4^{-}+16 \mathrm{H}^{+}+10 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O} \\ & 5 \mathrm{SO}_3^{2-}+5 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{SO}_4^{2-}+10 \mathrm{H}^{+}+10 e^{-} \end{aligned}$$

$2 \mathrm{MnO}_4^{-}+5 \mathrm{SO}_3^{2-}+6 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{SO}_4^{2-}+3 \mathrm{H}_2 \mathrm{O}$

This represents the correct balanced redox equation.

(d) Write the O. N. of all the atoms above their respective symbols.

(d) Write the O.N. of all the atoms above their respective symbols.

Divide skeleton equation into two half reactions

Reduction half reaction $\mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{2+}$

Oxidation half reaction $\mathrm{Br}^{-} \rightarrow \mathrm{Br}_2$

To balance reduction half reaction

$$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$$

To balance oxidation half reaction

$$2 \mathrm{Br}^{-} \longrightarrow \mathrm{Br}_2+2 e^{-}$$

To balance the reaction

$$\begin{aligned} 2 \mathrm{MnO}_4^{-}+16 \mathrm{H}^{+}+10 e^{-} & \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O} \\ 10 \mathrm{Br}^{-} & \longrightarrow 5 \mathrm{Br}_2+10 e^{-} \end{aligned}$$

$$2 \mathrm{MnO}_4^{-}+10 \mathrm{Br}^{-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{Br}_2+8 \mathrm{H}_2 \mathrm{O}$$

This represents the correct balanced ionic equation.

$$\mathrm{A.\to(4)\quad B.\to(5)\quad C.\to(3)\quad D.\to(1)}$$

Suppose that $x$ be the oxidation states of central atoms.

A. Oxidation number of Cr in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$

$$\begin{aligned} 2 x+7(-2) & =-2 \\ 2 x-14 & =-2 \\ 2 x & =+12 \\ x & =+6 \end{aligned}$$

$$\begin{aligned} &\text { B. Oxidation number of } \mathrm{Mn} \text { in } \mathrm{MnO}_4^{-}\\ &\begin{aligned} x+4(-2) & =-1 \\ x-8 & =-1 \\ x & =+7 \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { C. Oxidation number of } \mathrm{V} \text { in } \mathrm{VO}_3^{-}\\ &\begin{aligned} x+3(-2) & =-1 \\ x-6 & =-1 \\ x & =+5 \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { D. Oxidation number of } \mathrm{Fe} \text { in } \mathrm{FeF}_6^{3-}\\ &\begin{aligned} x+6(-1) & =-3 \\ x-6 & =-3 \\ \text { or }\quad x & =+3 \end{aligned}\\ \end{aligned}$$

A. $\rightarrow$ (5)

B. $\rightarrow$ (4)

C. $\rightarrow$ (3)

D. $\rightarrow$ (2)

E. $\rightarrow(6)$

A. lons having positive charge - Cation

B. The sum of oxidation number of all atoms in a neutral molecule - Zero

C. Oxidation number of hydrogen ion $\left(\mathrm{H}^{+}\right)-+1$

D. Oxidation number of fluorine in $\mathrm{NaF}--1$

E. lons having negative charge - Anion