Redox Reactions

The complete balance reaction is

$ 2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \longrightarrow 2 \mathrm{CuI}+\mathrm{I}_2 $

CuI further reacts to form $\mathrm{Cu}_2 \mathrm{I}_2$.

Moles of $\mathrm{Cu}^{2+}$

$ \begin{aligned} & =\text { Volume × concentration } \\ & =0.1 \mathrm{~L} \times 0.05=0.005 \mathrm{~mol} \end{aligned} $

Moles of $\mathrm{I}^{-}=\mathrm{IL} \times 0.1=0.1 \mathrm{~mol}$ 2 moles of $\mathrm{Cu}^{2+}$ react with 4 moles of $\mathrm{I}^{-}$.

Ratio of $\mathrm{Cu}^{2+}$ to $\mathrm{I}^{-}$required is $\frac{2}{4}=\frac{1}{2}$

Ratio of available $\mathrm{Cu}^{2+}$ to $\mathrm{I}^{-}=0.05$.

Since $0.05<0.5$. Thus, $\mathrm{Cu}^{2+}$ is limiting reactant.

From the stoichiometry, 2 moles of $\mathrm{Cu}^{2+}$ produces 1 moles of $\mathrm{I}_2$

$ \begin{aligned} \text { Moles of } \mathrm{I}_2 & =\frac{1}{2} \times \text { moles of } \mathrm{Cu}^{2+}=\frac{0.005}{2} \\ & =0.0025 \mathrm{~mol} \\ & =2.5 \times 10^{-3} \end{aligned} $

From the stoichiometry, 2 moles of $\mathrm{Cu}^{2+}$ produces 2 moles of CuI and 2 moles of CuI produces 1 mole of $\mathrm{Cu}_2 \mathrm{I}_2$.

$ \begin{aligned} \text { Moles of } \mathrm{Cu}_2 \mathrm{I}_2 & =\frac{0.005}{2}=0.0025 \mathrm{~mol} \\ & =2.5 \times 10^{-3} \end{aligned} $

The balanced chemical equation:

$ \begin{aligned} 2 \mathrm{Cr}(\mathrm{OH})_3 + \mathrm{IO}_3^{-} + 4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{CrO}_4^{2-} + \mathrm{I}^{-} + 5 \mathrm{H}_2 \mathrm{O} \end{aligned} $

Match the given general form to the balanced equation:

$ \begin{aligned} x \mathrm{Cr}(\mathrm{OH})_3 + y(\mathrm{IO}_3)^{-} + z(\mathrm{OH})^{-} \longrightarrow a(\mathrm{CrO}_4)^{2-} + b(\mathrm{I})^{-} + c(\mathrm{H}_2 \mathrm{O}) \end{aligned} $

Compare the numbers of each substance:

From the balanced equation, $x=2$, $y=1$, $z=4$, $a=2$, $b=1$, $c=5$

Check the answer:

Option (b) is wrong because $a + b = 2 + 1 = 3$, not 7.

When ammonium dichromate is thermally decomposed, it produces nitrogen gas, chromium (III) oxide and water vapour.

$ \begin{aligned} \left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7(s) \longrightarrow \mathrm{Cr}_2 \mathrm{O}_3(s)+ & \mathrm{N}_2(g) +4 \mathrm{H}_2 \mathrm{O}(g) \end{aligned} $

We need to find out where nitrogen has the lowest (most negative) oxidation state among the options. To do this, we will calculate the oxidation state of nitrogen in each compound.

(a) In $\mathrm{NH}_2\mathrm{OH}$ (Hydroxylamine):

Let nitrogen's oxidation state be $x$. Hydrogen (H) is +1 and oxygen (O) is -2.

There are 2 H atoms, so $2 \times (+1) = +2$. There is 1 O atom, so $(-2)$. There is also 1 extra H (in OH), so another $+1$.

$ \begin{array}{r} x+2(+1)+(-2)+(+1)=0 \\ x+1=0 \\ x=-1 \end{array} $

So, in hydroxylamine, nitrogen is at −1.

(b) In $\mathrm{NH}_4\mathrm{Cl}$ (Ammonium chloride):

This compound has an ammonium ion ($\mathrm{NH}_4^+$) and a chloride ion ($\mathrm{Cl}^-$).

Let nitrogen's oxidation state be $x$. Hydrogen is +1, chlorine is -1.

There are 4 H atoms, so $4 \times (+1) = +4$.

$ \begin{aligned} x+4 \times 1-1 & =0 \\ x+4(+1) & =+1 \\ x+4 & =1 \\ x & =-3 \end{aligned} $

So, in ammonium chloride, nitrogen is at −3.

(c) In $\mathrm{N}_2\mathrm{H}_4$ (Hydrazine):

Let nitrogen's oxidation state be $x$. Hydrogen is +1.

There are 2 nitrogen and 4 hydrogen atoms.

$ \begin{aligned} 2x+4(+1) & =0 \\ 2x & =-4 \\ x & =-2 \end{aligned} $

So, in hydrazine, nitrogen is at −2.

(d) In $\mathrm{HNO}_2$ (Nitrous acid):

Let nitrogen's oxidation state be $x$. Hydrogen is +1, oxygen is -2.

There is 1 H and 2 O.

$ \begin{array}{r} (+1)+x+2(-2)=0 \\ 1+x-4=0 \\ x-3=0 \\ x=+3 \end{array} $

So, in nitrous acid, nitrogen is at +3.

Now, let's compare the results: $-1,\ -3,\ -2,\ +3$. The smallest value is −3. This happens in $\mathrm{NH}_4\mathrm{Cl}$ (ammonium chloride).

The number of moles of oxalate ions oxidised by one mole of permanganate ions in acidic medium is 2.5. The complete reaction involved is as follows.

$ \begin{aligned} \frac{5}{2} \mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{MnO}_4^{-} & +8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}+5 \mathrm{CO}_2 \end{aligned} $

The reactions that happen are:

$ \begin{aligned} \mathrm{MnO}_4^{-}+5 e^{-} & \longrightarrow \mathrm{Mn}^{2+} \quad \text{(1 mole of } \mathrm{MnO}_4^{-} \text{ takes 5 electrons)} \\ \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e^{-} & \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \quad \text{(1 mole of } \mathrm{Cr}_2 \mathrm{O}_7^{2-} \text{ takes 6 electrons)} \\ \mathrm{Fe}^{2+} & \longrightarrow \mathrm{Fe}^{3+} \quad \text{(Each Fe}^{2+} \text{ loses 1 electron.)} \end{aligned} $

Step 1: How many Fe²⁺ ions does 1 mole of MnO₄^- oxidize?

1 mole of MnO₄^- can take in 5 electrons. Each Fe²⁺ gives up 1 electron to get oxidized. So, 1 mole of MnO₄^- can oxidize 5 moles of Fe²⁺. This number is called $x$.

Step 2: How many Fe²⁺ ions does 1 mole of Cr₂O₇^2- oxidize?

1 mole of Cr₂O₇^2- can take in 6 electrons. That means it can oxidize 6 moles of Fe²⁺.

Step 3: What is the answer in terms of $x$?

Since $x = 5$, and Cr₂O₇^2- can oxidize 6 moles, the answer is $\frac{6x}{5}$ moles of Fe²⁺ ions.

In acidic medium,

$ \begin{aligned} 5 \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+} \longrightarrow & \mathrm{Mn}^{2+} + 4 \mathrm{H}_2 \mathrm{O}+5 \mathrm{Fe}^{3+} \end{aligned} $

1 mole of $\mathrm{MnO}_4^{-}$requires 5 moles of $\mathrm{Fe}^{2+} x=5$

$ \begin{aligned} \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \longrightarrow 2 \mathrm{Cr}^{3+} & +7 \mathrm{H}_2 \mathrm{O} +6 \mathrm{Fe}^{3+} \end{aligned} $

1 mole of $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ are needed to give 6 moles of $\mathrm{Fe}^{3+}, y=6$.

$ x+y=5+6=11 $

The balanced chemical reaction is:

$ \mathrm{C} + 2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{CO}_{2} + 2 \mathrm{SO}_{2} + 2 \mathrm{H}_{2} \mathrm{O} $

In this reaction:

Product $ X $ is Carbon dioxide $(\mathrm{CO}_{2})$

Product $ Y $ is Sulphur dioxide $(\mathrm{SO}_{2})$

Given reaction is,

In this reaction,

Cl of $ \mathrm{KClO}_{3} $ (oxidation number $ =+5 $ ) is reduced to $ \mathrm{Cl}^{-} $and O of $ \mathrm{KClO}_{3} $

$ \mathrm{KClO}_{3} $ (oxidation number $ =-2 $ ) is oxidised to $ \mathrm{O}_{2} $.

The correct match is A-III, B-I, C-IV, D-II.

For $ \mathrm{Na}_{2} \mathrm{CO}_{3} $ the equivalent weight is $ \frac{M}{2} $ as it is diprotic base.

In acidic medium, the $ n $-factor for $ \mathrm{KMnO}_{4} $ is 5 . So, its equivalent weight is $ \frac{M}{5} $.

In neutral medium, the $ n $-factor for $ \mathrm{K}_{2} 2_{2} \mathrm{r}_{2} $ is 6 . So, its equivalent weight is $ \frac{M}{6} $. In neutural medium, the $ n $-factor for $ \mathrm{KMnO}_{4} $ is 3. So, its equivalent weight is $ \frac{M}{3} $.

20 Volume of $ \mathrm{H}_{2} \mathrm{O}_{2} $ solution refers to 1 L of $ \mathrm{H}_{2} \mathrm{O}_{2} $ solution that will liberale 20 L of $ \mathrm{O}_{2} $ at NTP.

$ \underset{68 \mathrm{~g}}{2 \mathrm{H}_{2} \mathrm{O}_{2}} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\underset{22.4 \mathrm{~L} \text { at NTP }}{\mathrm{O}_{2}} $

22.4 L of $ \mathrm{O}_{2} $ at NTP produce from $ \mathrm{H}_{2} \mathrm{O}_{2}=68 \mathrm{~g} $

20 L of $ \mathrm{O}_{2} $ at NTP produce from $ \mathrm{H}_{2} \mathrm{O}_{2} $

$ =\frac{68}{22.4} \times 20=60.71 \mathrm{~g} $

Strength of $ \mathrm{H}_{2} \mathrm{O}_{2}=60.71 \mathrm{~g} / \mathrm{L} $

Equivalent weight of $ \mathrm{H}_{2} \mathrm{O}_{2}=\frac{68}{32} \times 8=17 $

$ \text { Normality }=\frac{\text { Strength }}{\text { Equivalent weight }}=\frac{60.71}{17} $

Normality $ \approx 3.570 \mathrm{~N} $

In the given three oxoanions, the oxidation state of $\mathrm{V}, \mathrm{Cr}$ and Mn in $\mathrm{VO}_2^{+}$, $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ and $\mathrm{MnO}_4^{-}$are $+5,+6$ and +7 respectively.

Hence, the correct increasing order is, $\mathrm{VO}_2^{+}<\mathrm{Cr}_2 \mathrm{O}_7^{2-}<\mathrm{MnO}_4^{-}$

Sodium is oxldised from 0 to +1. Water is reduced from +1 to 0.

Fluorine is reduced from 0 to -1. Water is oxidised from -2 to 0.

The structure of $\mathrm{C}_3 \mathrm{O}_2$ is

$ \mathrm{O}=\mathrm{C}=\mathrm{C}=\mathrm{C}=\mathrm{O} $

Assigning the oxidation number to each C -atom in molecule considering oxidation number of oxygen as -2

We get, oxidation state as

$ \mathrm{O}=\stackrel{+2}{\mathrm{C}}=\stackrel{0}{\mathrm{C}}=\stackrel{+2}{\mathrm{C}}=\mathrm{O} $

Thus, two C -atoms linked with O -atom has +2 oxidation state and carbon atom in the centre has 0 oxidation state.

The reaction (a) produce nitrogen

(II) oxide as one of the product.

$ \begin{aligned} 3 \mathrm{Cu}+8 \mathrm{HNO}_3 \text { dil. } \longrightarrow & 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2 +2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O} \end{aligned} $

Other reactions are as follows :

$ \begin{aligned} & \mathrm{Cu}+4 \mathrm{HNO}_3 \text { (conc.) } \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2 +2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O} \\ & 4 \mathrm{Zn}+10 \mathrm{HNO}_3 \text { (dil.) } \rightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_2 +5 \mathrm{H}_2 \mathrm{O}+\mathrm{N}_2 \mathrm{O} \\ & \mathrm{Zn}+4 \mathrm{HNO}_3 \text { (conc.) } \rightarrow \mathrm{Zn}\left(\mathrm{NO}_3\right)_2 +2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{NO}_2 \end{aligned} $

$\mathrm{SO}_2$ is the strongest reducing agent. In $\mathrm{SO}_2$, sulphur have +4 oxidation state and it will attain +6 oxidation state and ion $\mathrm{SO}_4^{2-}$. That means it undergoes oxidation (going from lower oxidation state to higher oxidation state). $\mathrm{So}, \mathrm{SO}_2$ will oxidises itself and reduces others and act as reducing agent, while $\mathrm{TeO}_2, \mathrm{TeO}_3$ and $\mathrm{SO}_3$ acts as an oxidising agent.

In above reaction, oxidation and reduction takes place simultaneously that means it is a redox reaction but it is not a disproportionation reaction because same element is not simultaneously oxidised and reduced.

To determine which compound has the maximum equivalent weight, let's calculate the equivalent weights for each:

For $ \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O} $:

The formula for calculating equivalent weight is:

$ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} $

Here, n-factor is 2 because it involves a two-electron transfer process. The molar mass of $ \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O} $ is:

$ 2(23) + 2(12) + 4(16) + 2(18) = 170 $

$ \text{Equivalent weight} = \frac{170}{2} = 85 \, \mathrm{g} \, \mathrm{eq}^{-1} $

For $ \mathrm{KMnO}_4 / \mathrm{H}^{+} $:

This reaction has an n-factor of 5 due to the change in oxidation state from +7 to +2 for Mn. The molar mass is:

$ 39 + 55 + 4(16) = 158 $

$ \text{Equivalent weight} = \frac{158}{5} = 31.6 \, \mathrm{g} \, \mathrm{eq}^{-1} $

For $ \mathrm{KMnO}_4 / \mathrm{H}_2 \mathrm{O} $:

When Mn changes from +7 to +4, the n-factor is 3. The molar mass calculation remains the same:

$ 39 + 55 + 64 = 158 $

$ \text{Equivalent weight} = \frac{158}{3} = 52.6 \, \mathrm{g} \, \mathrm{eq}^{-1} $

For $ \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 / \mathrm{H}^{+} $:

The n-factor is 6 because chromium changes from an oxidation state of +6 to +3. The molar mass is:

$ 2(39) + 2(52) + 7(16) = 294 $

$ \text{Equivalent weight} = \frac{294}{6} = 49 \, \mathrm{g} \, \mathrm{eq}^{-1} $

The compound $ \mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot 2 \mathrm{H}_2 \mathrm{O} $ has the highest equivalent weight of 85 g eq$^{-1}$.

Disproportionation reactions are the reactions in which an element undergoes oxidation and reduction simultaneously forming two respective products. (One with higher oxidation state and one with lower.)

$\therefore \mathrm{A}, \mathrm{B}, \mathrm{D}$ only are disproportionation reactions.

Reaction of ferrous oxalate oxidised by $\mathrm{KMnO}_4$ is

Number of gram equivalent of $\mathrm{KMnO}_4$

$=$ Number of gram equivalent of ferrous oxalate

Number of moles $\times n$-factor

$=$ number of moles $\times n$-factor

Number of moles of $\mathrm{KMnO}_4 \times 2=1 \times 5$

Number of moles of $\mathrm{KMnO}_4=5 / 2$

Disproportionation reactions are those in which one element undergoes both oxidation and reduction both simultaneously.

In the reaction,

Oxygen in reactant $\mathrm{H}_2 \mathrm{O}_2$ is in -1 oxidation state. Oxygen in product $\mathrm{H}_2 \mathrm{O}$ is in -2 oxidation state and in other product $\mathrm{O}_2$, it is in 0 oxidation state. So, in this reaction oxygen undergoes oxidation and reduction both simultaneously.

In the reaction,

Chlorine in reactant $\mathrm{Cl}_2$ is in 0 oxidation state. Chlorine in product NaCl is in -1 oxidation state and in other product $\mathrm{NaOCl}, \mathrm{Na}$ is in +1 oxidation state, 0 is in -2 oxidation state and Cl is in +1 oxidation state. So, in this reaction chlorine undergoes oxidation and reduction both simultaneously.

$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ mostly react in acidic medium as oxidising agent. The reaction occurs as follows:

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

∵ Reaction involve $6 e^{-}$electrons. Thus, Equivalent weight

$ \begin{aligned} & =\frac{\text { Molar mass }}{\text { Number of electrons }}=\frac{294}{6} \\ & =49 \end{aligned} $

∵ We have to prepare 0.04 N of 250 mL solution.

$ (\because 250 \times 4=1000 \mathrm{~mL}) $

Thus, $\frac{0.04}{4}=0.01$

$ \text { and weight of } \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \text { is }=49 \times 0.01 $

                                                   $ =0.49 \mathrm{~g} $

Hence, option (d) is the correct answer.

$\because$ In acidic medium, $\mathrm{KMnO}_4$, i.e.

$\mathrm{MnO}_4^{-}$changes to $\mathrm{Mn}^{2+}$ as follows :

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

Hence, equivalent weight $=\frac{158}{5}$

(Given, molecular weight of $\mathrm{KMnO}_4=158 \mathrm{~g}$ and 5 electrons are used)

$ \therefore \quad \frac{158}{5}=31.6 \mathrm{~g} $

Hence, option (c) is the correct answer.

The reaction will faster in aqueous HCl because chlorine ion present in HCl get oxidised into chlorine gas by $\mathrm{KMnO}_4$, so the amount of $\mathrm{KMnO}_4$ will be more as it oxidises both oxalic acid and chlorine ion when HCl is present.