Redox Reactions

Moles of $\mathrm{Cu}^{2+}: n_{\mathrm{Cu}^{2+}}=M_{\mathrm{Cu}^{2+}} \times V_{\mathrm{Cu}^{2+}}$

$ =0.05 \times 0.1=0.005 \mathrm{~mol} $

For the reaction,

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

2 moles of $\mathrm{Cu}^{2+}$ produces 1 mol of $\mathrm{I}_2$

Moles of $\mathrm{I}_2: n_{1_2}=\frac{1}{2} \times n_{\mathrm{Cu}^{2+}}=0.0025 \mathrm{~mol}$

For the reaction;

$ \mathrm{I}_2+2 \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \longrightarrow 2 \mathrm{NaI}+\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6 $

1 mole $\mathrm{I}_2$ react with 2 moles of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$

$ n_{\mathrm{Na} S_2 S_3}=2 \times 0.0025=0.005 \mathrm{~mol} $

Volume of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3: V_{\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3}=\frac{n_{\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3}}{M_{\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3}}$

$ =\frac{0.005}{0.01}=0.5 \mathrm{~L} \text { or } 500 \mathrm{~mL} $

Statements given in II and III are correct.

The correct balanced equation is,

$ \begin{gathered} \mathrm{P}_4+3 \mathrm{OH}^{-}+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3+3 \mathrm{H}_2 \mathrm{PO}_2^{-} \\ a=1, b=3, c=3, d=1, e=3 \\ a+b+c=7, b+c-e=3 \end{gathered} $

Balanced oxidation half reaction

$ \mathrm{H}_2 \mathrm{~S} \rightarrow \mathrm{~S}+2 \mathrm{H}^{+}+2 e^{-} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Balanced reduction half reaction

$ \begin{aligned} \mathrm{MnO}_4^{-} \rightarrow 8 \mathrm{H}^{+}+5 e^{-} \longrightarrow & \mathrm{Mn}^{2+} + 4 \mathrm{H}_2 \mathrm{O} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Eq. (i) multiply by 5 and Eq. (ii) multiply by 2 then add

$ \begin{aligned} 2 \mathrm{MnO}_4^{-}+5 \mathrm{H}_2 \mathrm{~S}+6 \mathrm{H}^{+} \longrightarrow & 2 \mathrm{Mn}^{2+} + 5 \mathrm{~S}+8 \mathrm{H}_2 \mathrm{O} \end{aligned} $

5 mole $\mathrm{H}_2 \mathrm{~S}$ react with 2 mol of $\mathrm{MnO}_4^{-}$

$1 \mathrm{~mol} \mathrm{H}_2 \mathrm{~S}=\frac{2}{5} \mathrm{~mol} \mathrm{MnO}_4^{-}$or 0.4

In acidic condition, $\mathrm{KMnO}_4$ (Mn $=+7$ oxidation state) reduced by $\mathrm{H}_2 \mathrm{O}_2$ to $\mathrm{Mn}^{2+}$

So, $x=\mathrm{Mn}^{2+}$

In basic condition $\mathrm{KMnO}_4(\mathrm{Mn}=+7$ oxidation state)

is reduced by $\mathrm{H}_2 \mathrm{O}_2$ to $\mathrm{MnO}_2 y=\mathrm{MnO}_2$

Acidic $2 \mathrm{KMnO}_4+5 \mathrm{H}_2 \mathrm{O}_2+3 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow2 \mathrm{MnSO}_4+\mathrm{K}_2 \mathrm{SO}_4+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}_2 $

Basic

$ \begin{array}{r} 2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{MnO}_2+2 \mathrm{KOH} +2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{O}_2 \end{array} $

Among the given reactions, the reaction

$ \mathrm{I}_2(s)+2 \mathrm{KBr}(a q) \longrightarrow 2 \mathrm{KI}(a q)+\mathrm{Br}_2(l) $

is not feasible because iodine is less reactive than bromine. In a displacement reaction, more reactive element can displace a less reactive element in compound.

$\mathrm{MnSO}_4$ acts as the autocatalyst during the titration of $\mathrm{KMnO}_4$ and oxalic acid.

An autocatalyst is the substance that is formed during a chemical reaction and then speeds up the reaction.

Oxidation state of Mn in acidic medium

$\mathrm{KMnO}_4$ reduced to $\mathrm{Mn}^{2+}$

Thus, oxidation state $=+2$

In basic medium,

$\mathrm{KMnO}_4$ reduced to $\mathrm{MnO}_2$

$ 0=x+2(-2) \Rightarrow x=+4 $

Normality of $\mathrm{KMnO}_4$ solution,

$ N_{\mathrm{KMnO}_4}=n_{\text {factor }} \times M=0.4 \times 5=2 \mathrm{~N} $

Normality of $\mathrm{H}_2 \mathrm{O}_2$

$ N_{\mathrm{H}_2 \mathrm{O}_2}=\frac{\text { Volume strength }}{5.6}=\frac{10}{5.6} \mathrm{~N} $

Using normality equation

$ \begin{aligned} & N_1 V_1=N_2 V_2 \\ & N_{\mathrm{KMnO}_4} \times V_{\mathrm{KMnO}_4}=N_{\mathrm{H}_2 \mathrm{O}_2} \times V_{\mathrm{H}_2 \mathrm{O}_2} \\ & 2 \times 200= \frac{10}{5.6} \times V_{\mathrm{H}_2 \mathrm{O}_2} \\ & V_{\mathrm{H}_2 \mathrm{O}_2}= 224 \mathrm{~mL} \end{aligned} $

The following reactions are redox reactions.

Reaction (iii) is a double displacement reaction.

The process of acidifying chromate involves the following reaction:

$ 2 \mathrm{Na}_2 \mathrm{CrO}_4 + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 + \mathrm{H}_2 \mathrm{O} + \mathrm{Na}_2 \mathrm{O}_4 $

This can be simplified to:

$ \mathrm{CrO}_4^{2-} \xrightarrow{\mathrm{H}^{+}} \mathrm{Cr}_2 \mathrm{O}_7^{2-} $

In this reaction, chromate ions ($\mathrm{CrO}_4^{2-}$) are converted into dichromate ions ($\mathrm{Cr}_2 \mathrm{O}_7^{2-}$).

In the compound $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$, sodium ($\mathrm{Na}$) is in the +1 oxidation state and oxygen ($\mathrm{O}$) is in the -2 oxidation state. To find the oxidation state of chromium ($\mathrm{Cr}$) in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$, we analyze the overall charge balance:

Total charge contributed by two sodium atoms: $2 \times (+1) = +2$.

Each oxygen atom contributes a -2 charge, and there are 7 oxygen atoms: $7 \times (-2) = -14$.

The net charge on the ion $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ is -2.

Setting up the equation for chromium:

$ 2 + 2x - 14 = 0 $

Solving for $x$ gives:

$ 2x = 12 \\ x = +6 $

Thus, the oxidation state of chromium in $\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is +6.

Given, 100 mL of $0.1 \mathrm{M} \mathrm{Fe}^{2+}$ titrated with $\frac{1}{60} \mathrm{M} \mathrm{Cr}_2 \mathrm{O}_7^{2-}$ solution.

$ \text { The complete reaction is, } $

$ \begin{aligned} & \Rightarrow \quad 0.1 \mathrm{M} \quad \frac{1}{60} M =0.1 \mathrm{~N}=\frac{1}{10} \mathrm{~N}[\because N=M \times n \text {-factor }] \\ & \Rightarrow \quad V \mathrm{~mL} \quad 100 \mathrm{~mL}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(say) \\ & \ \\ & \Rightarrow \quad V=\frac{1}{0.1} \times\left(100 \times \frac{1}{10}\right)=100 \mathrm{~mL} \end{aligned} $

So, $\quad V=0.1 \mathrm{~L}$

When hydrogen sulfide ($ \mathrm{H}_2 \mathrm{S} $) is passed through acidic potassium permanganate ($ \mathrm{KMnO}_4 $), sulfur is produced.

The balanced chemical equation for this reaction is:

$ \mathrm{KMnO}_4 + \mathrm{H}_2 \mathrm{S} + \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{H}_2 \mathrm{O} + \mathrm{K}_2 \mathrm{SO}_4 + \mathrm{MnSO}_4 + \mathrm{S} $

In this reaction, $ \mathrm{KMnO}_4 $ acts as an oxidizing agent, converting $ \mathrm{H}_2 \mathrm{S} $ into sulfur ($ \mathrm{S} $).

In a neutral medium, the permanganate ion oxidizes iodine to form iodate. Here's the reaction illustrating this process:

$ \begin{aligned} & \mathrm{KMnO}_4 \rightleftharpoons \mathrm{MnO}_4^{-}+\mathrm{K}^{+} \\ & 2 \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O}+\mathrm{I}^{-} \longrightarrow \\ & 2 \mathrm{MnO}_2+2 \mathrm{OH}^{-}+\mathrm{IO}_3^{-} \end{aligned} $

This reaction shows how permanganate ions in a neutral solution can facilitate the conversion of iodide ions to iodate ions, with manganese dioxide and hydroxide ions as additional products.

The complete reaction is as follows,

(a) $\mathrm{S}+2 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow 3 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$

Here, sulphur $(\mathrm{S})$ is oxidised to $\mathrm{SO}_2$

$\mathrm{H}_2 \mathrm{SO}_4$ is reduced to $\mathrm{SO}_2$.

Hence, both $X$ and $Y$ are $\mathrm{SO}_2$.

Let's analyze each statement step by step:

For Statement I (Decomposition of potassium chlorate):

Consider the reaction:

$2KClO_3 \rightarrow 2KCl + 3O_2$

In potassium chlorate (KClO₃), let's determine the oxidation state of chlorine.

For KClO₃, potassium (K) has an oxidation number of +1 and oxygen (O) is typically -2. Setting chlorine's oxidation number as x:

$+1 + x + 3(-2) = 0 \quad \Rightarrow \quad x = +5$

In potassium chloride (KCl), K is +1 and Cl must be -1 (to balance the compound).

Since chlorine goes from +5 to -1, it has gained electrons. This is a reduction. So Statement I is correct.

For Statement II (Reaction of sodium with oxygen to form Na₂O):

Consider the reaction:

$4Na + O_2 \rightarrow 2Na_2O$

Sodium (Na) starts with an oxidation state of 0 and in Na₂O, sodium is +1.

Oxygen (O₂) starts with 0 and in Na₂O, it is -2.

With sodium being oxidized (gain in oxidation number) and oxygen being reduced (loss in oxidation number), this clearly is a redox reaction. So Statement II is correct.

Since both statements are correct, the correct option is:

Option A

Both statements I and II are correct.

In a neutral medium, potassium permanganate oxidizes $ I^{-} $ to $ \text{IO}_3^{-} $ (iodate). The balanced chemical reaction for this process is as follows:

$ 2 \text{KMnO}_4 + \text{H}_2\text{O} + \text{KI} \longrightarrow 2 \text{MnO}_2 + 2 \text{KOH} + \text{KIO}_3 $

In this reaction, the permanganate ion acts as an oxidizing agent, facilitating the conversion of iodide ions into iodate ions.

When an atom is bonded to similar atoms, has zero oxidation state.

$\therefore$ Two middle sulphur atoms have zero oxidation state.

For other two sulphur atoms.

$\begin{aligned} 2 x+6 \times(-2) & =-2 \\ 2 x-12 & =-2 \\ 2 x & =-2+12 \\ 2 x & =+10 \\ x & =+5 \end{aligned}$

$\therefore$ The oxidation number on S-atoms in $\mathrm{S}_4 \mathrm{O}_6^{2-}$ is $+5,0,0+5$.

$\mathrm{Mg}+\mathrm{NO}_3^{-} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{NH}_3$

The complete balanced equation of the given reaction is

$4 \mathrm{Mg}+\mathrm{NO}_3^{-}+9 \mathrm{H}^{\oplus} \longrightarrow 4 \mathrm{Mg}^{2+}+\mathrm{NH}_3+3 \mathrm{H}_2 \mathrm{O}$

Given, conc. of $\mathrm{NO}_3^{-}=0.1 \mathrm{~M}$

Volume of $\mathrm{NO}_3^{-}=100 \mathrm{~mL}$

We know that,

$\text { Molarity }=\frac{\text { Number of moles }}{\text { Volume of solution (in } \mathrm{L} \text { ) }}$

$\begin{aligned} & \text { Hence, number of moles }=\text { molarity } \times \text { volume of } \\ & \text { solution (in } \mathrm{L}) \\ & \begin{aligned} \text { So, number of moles } & =0.1 \times 100 \times 10^{-3} \\ & =10 \mathrm{~m} \mathrm{~mol} \\ & =10^{-2} \mathrm{~mol} \end{aligned} \end{aligned}$

From above Eq., 1 mole $\mathrm{NO}_3^{-}$ reacts with 4 moles Mg. So, $10^{-2}$ mole $\mathrm{NO}_3^{-}$ reacts with $4 \times 10^{-2}$ moles Mg. Now, to calculate the mass of Mg required, use the formula,

$\text { Number of moles }=\frac{\text { Given mass }}{\text { Molar mass }}$

Molar mass of $\mathrm{Mg}=24 \mathrm{~g} \mathrm{~mol}^{-1}$

On substituting the value,

$4 \times 10^{-2}=\frac{\text { Mass of } \mathrm{Mg}}{24}$

We will get,

Mass of $\mathrm{Mg}=24 \times 4 \times 10^{-2}$

Mass of $\mathrm{Mg}=0.96 \mathrm{~g}$

The given reaction is:

$\mathrm{SO}_3^{2-}(aq) + \mathrm{Br}_2(l) + \mathrm{H}_2O(l) \rightarrow \mathrm{SO}_4^{2-}(aq) + 2 \mathrm{Br}^-(aq) + 2 \mathrm{H}^+$

To determine the oxidation state of sulfur (S) in the product, we first identify that the sulfur-containing product is $\mathrm{SO}_4^{2-}$ (sulfate ion). Let's denote the oxidation state of sulfur as $ x $.

For the sulfate ion $\mathrm{SO}_4^{2-}$:

• Each oxygen (O) atom has an oxidation state of $-2$.


• There are four oxygen atoms, contributing a total of $ 4 \times (-2) = -8 $.

The net charge on the sulfate ion is $-2$. Using these data, we set up the following equation to solve for $ x $:

$ x + 4(-2) = -2 $

Simplifying this:

$ x - 8 = -2 $

Solving for $ x $:

$ x = -2 + 8 $

$ x = +6 $

Therefore, the oxidation state of sulfur in the sulfate ion $ \mathrm{SO}_4^{2-} $ is $ +6 $.

Phosphorus reacts with $\mathrm{HNO}_3$ to give $\mathrm{H}_3 \mathrm{PO}_4$ and $\mathrm{NO}_2$. In this reaction, phosphorus is oxidised to $\mathrm{H}_3 \mathrm{PO}_4$ and also N is reduced to $\mathrm{NO}_2$. The reaction is

$\mathrm{P}+5 \mathrm{HNO}_3 \longrightarrow \mathrm{H}_3 \mathrm{PO}_4+5 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O} $

Structure of Br$_3$O$_8$ is

Each terminal Br has three oxygen atoms attached to it. So, the oxidation state of Br -atom on terminals is +6 .

The bromine atom in the middle has two oxygen atoms attached to it. So, the oxidation state of Br atom in middle is +4.

Hence, the oxidation states of three Br -atoms in $\mathrm{Br}_3 \mathrm{O}_8$ molecule are $+6,+4,+6$.

(a) $\mathrm{Tl}$ is more stable in +1 oxidation state, then +3 oxidation state, therefore $\mathrm{Tl}^{3+}$ salts are oxidising agents.

(b) $\mathrm{Ga}^{+}-2 e^{-} \rightarrow \mathrm{Ga}^{3+}$ (More stable)

$\therefore \mathrm{Ga}^{+}$ is a reducing agent.

(c) $\mathrm{Pb}^{4+}+2 e^{-} \rightarrow \mathrm{Pb}^{2+}$ (More stable)

$\mathrm{Pb}^{4+}$ is an oxidising agent.

(d) $\mathrm{As}^{5+}+2 e^{-} \rightarrow \mathrm{As}^{3+}$ (Less stable)

Thus, $\mathrm{As}^{5+}$ cannot be reduced so as${ }^{5+}$ salts are not a good oxidising agent. An oxidising agent is defined as substance whose oxidation number decreases while reducing agent defined as substance whose oxidation number increases.

Disproportionation is a specific type of redox reaction in which a species is simultaneously reduced and oxidised to form two different products.

Electronic configuration of Cl is $1s^2$ 2$s^2$ 2$p^2$ 3$s^2$ 3$p^6$ 4$s^1$. It can exhibit maximum oxidation state of +7.

All species can exhibit disproportionation reaction which simultaneously oxidise as well as reduce.

In ClO$_4^-$, Cl cannot undergo oxidation as +7 is maximum oxidation state exhibited by chlorine.

The unbalanced redox reaction is

$\mathrm{MnO}_4^{-}+\mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$

(i) Balanced $\mathrm{C}$ atoms

$\stackrel{+7}{\mathrm{MnO}_4^{-}}+\mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$

Oxidation state of Mn decreases from +7 to +2 .

To balance the increase or decrease in oxidation number, multiply Mn containing species with 2 and $\mathrm{C}$ containing species with 5 .

$2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2$

(ii) Balance '$\mathrm{O}$' atoms by adding 8 water molecules to the product side.

$2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}$

(iii) Balance '$\mathrm{H}$' atoms by adding $16 \mathrm{H}^{+}$ ions to the reactant side.

$\begin{aligned} 2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \rightarrow \\ 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O} \end{aligned}$

Hence, $\mathrm{MnO}_4^{-} \Rightarrow 2, \mathrm{C}_2 \mathrm{O}_4^{2-} \Rightarrow 5 ; \mathrm{H}^{+}=16$

When the old oil painting are kept in the atmosphere for a longer time the tracer of gas present in atmosphere converts lead oxide (PbO) into PbS which is black in colour.

Therefore, these paintings get transhed. It can be restored by keeping then in H$_2$O$_2$ solution for sometime as a result of which lead sulphide is oxidised to lead sulphate.

So, A is true but R is false.

$\mathrm{KMn{O_4}\buildrel {} \over \longrightarrow {K^ + } + \mathop {MnO_4^ - }\limits_{( + 7)}}$

In presence of a reducing agent, $\mathrm{MnO}_4^{-}$ acts as a self-indicator its and end point is pink to colourless. This is because in $\mathrm{MnO}_4^{-}, \mathrm{Mn}$ is in +7 oxidation state and is in highest oxidation state. Thus, will tends to get reduced and easily takes electrons and being a charge transfer (CT) complex, shows intense colour.

After titration, it is MnO$_2$ in +4 state and has no electronic charge transfer, therefore colourless.