Redox Reactions

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.

$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ 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.

The reaction between sodium hydroxide (NaOH) and oxalic acid (H₂C₂O₄) is as follows :

$ \text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O} $

We can see that 1 mole of oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) reacts with 2 moles of sodium hydroxide (NaOH).

Given that 50 mL of 0.5 M oxalic acid is needed to neutralize the sodium hydroxide, we can find the number of moles of oxalic acid that reacted :

$ \text{Moles of H}_2\text{C}_2\text{O}_4 = \text{Molarity} \times \text{Volume (L)} = 0.5 \, \text{mol/L} \times 0.05 \, \text{L} = 0.025 \, \text{mol} $

Since 1 mole of $\text{H}_2\text{C}_2\text{O}_4$ reacts with 2 moles of $\text{NaOH}$, the number of moles of $\text{NaOH}$ in 25 mL solution would be twice that of $\text{H}_2\text{C}_2\text{O}_4$ :

$ \text{Moles of NaOH} = 0.025 \, \text{mol} \times 2 = 0.05 \, \text{mol} $

Now, to find the mass of NaOH, we multiply the number of moles by the molar mass of NaOH (40 g/mol) :

$ \text{Mass of NaOH} = \text{Number of moles} \times \text{Molar mass} = 0.05 \, \text{mol} \times 40 \, \text{g/mol} = 2 \, \text{g} $

However, the question asks for the amount of $\text{NaOH}$ in 50 mL of the given solution. Since we've found the amount in 25 mL, we just need to double our result to find the amount in 50 mL :

$ 2 \, \text{g} \times 2 = 4 \, \text{g} $

So, the correct answer is Option B : 4 g.

The balanced reaction of oxalate with permanganate in acidic medium is :

$5 \, \text{C}_2\text{O}_4^{2-} + 2 \, \text{MnO}_4^- + 16 \, \text{H}^+ \rightarrow 10 \, \text{CO}_2 + 2 \, \text{Mn}^{2+} + 8 \, \text{H}_2\text{O}$

In this redox reaction, the oxalate ion $\text{C}_2\text{O}_4^{2-}$ is oxidized to carbon dioxide $\text{CO}_2$, and the permanganate ion $\text{MnO}_4^-$ is reduced to manganese(II) ion $\text{Mn}^{2+}$.

The oxidation half-reaction, representing the change for the oxalate ion, is :

$\text{C}_2\text{O}_4^{2-} \rightarrow 2 \, \text{CO}_2 + 2 \, e^-$

From this half-reaction, it's clear that each oxalate ion produces two CO₂ molecules and releases two electrons in the process.

So, the number of electrons involved in producing one molecule of CO₂ is :

$\frac{2 \, e^-}{2 \, \text{CO}_2} = 1 \, e^-/\text{CO}_2$

Therefore, the correct answer is Option C: One electron is involved in the production of one molecule of CO₂.

Let oxidation states of phosphorus in H₃PO₂, H₃PO₄, H₃PO₃ and H₄P₂O₆ be p, q, r and s respectively.

Oxidation state of hydrogen = + 1

Oxidation state of oxygen = $-$2

Thus, in H₃PO₂ :

3 $\times$ (+1) + p + 2 $\times$ ($-$2) = 0 $\therefore$ p = +1

In H₃PO₄ :

3 $\times$ (+1) + q + 4 $\times$ ($-$2) = 0 $\therefore$ q = +5

In H₃PO₃ :

3 $\times$ (+1) + r + 3 $\times$ ($-$2) = 0 $\therefore$ r = + 3

In H₄P₂O₆ :

4 $\times$ (+1) + 2s + 6 $\times$ ($-$2) = 0 $\therefore$ s = +4

Thus, the order of oxidation state is :

H₃PO₄ > H₄P₂O₆ > H₃PO₃ > H₃PO₂

Fe³⁺ is reduced to Fe²⁺ by H₂O₂/NaOH and Na₂O₂/H₂O.

$2F{e^{3 + }} + {H_2}{O_2} + O{H^ - } \to 2F{e^{2 + }} + 2{H_2}O + {O_2}$

$N{a_2}{O_2} + {H_2}O \to {H_2}{O_2} + NaOH$

In an aqueous solution, the reduction of the metal center in permanganate ions ($MnO_4^-$) varies depending on the medium :

Acidic Medium : Permanganate acts as a strong oxidizing agent. The reaction can be represented as :

$8H^+ + 5e^- + MnO_4^- \to Mn^{2+} + 4H_2O$

This indicates that 5 electrons are involved in the reduction process in an acidic medium.

Neutral Medium : Permanganate acts as a moderate oxidizing agent. The reaction in a neutral medium is :

$2H_2O + 3e^- + MnO_4^- \to MnO_2 + 4OH^-$

Here, 3 electrons are involved in the reduction process in a neutral medium.

Alkaline Medium : In alkaline conditions, the reaction is similar to that in a neutral medium :

$2H_2O + 3e^- + MnO_4^- \to MnO_2 + 4OH^-$

Again, 3 electrons are involved in the reduction process in an alkaline medium.

When the same molecule is oxidised or reduced, then the reaction is called disproportionation reaction.

The reaction in which oxidation and reduction takes place simultaneously are called redox reactions. Thus,

(A)

(B)

(C)

(D)

The given reaction is:

$3ClO^- (aq) \to ClO_3^- (aq) + 2Cl^- (aq)$

This reaction is an example of a Disproportionation Reaction. Disproportionation reactions are a type of redox reaction in which the same element is simultaneously oxidized and reduced. In this given reaction, chlorine undergoes both oxidation and reduction.

To understand why it is a disproportionation reaction, let's assign oxidation states to chlorine in different species:

• In $ClO^-$, chlorine has an oxidation state of +1.


• In $ClO_3^-$, chlorine has an oxidation state of +5.


• In $Cl^-$, chlorine has an oxidation state of -1.

In going from $ClO^-$ to $ClO_3^-$, the oxidation state of chlorine increases from +1 to +5, indicating oxidation. Simultaneously, in going from $ClO^-$ to $Cl^-$, the oxidation state of chlorine decreases from +1 to -1, indicating reduction. Since one species of chlorine is getting oxidized while another is getting reduced, and both processes involve the same element (chlorine in this case), it classifies as a disproportionation reaction.

Therefore, the correct answer is:

Option C: Disproportionate reaction

To identify the species with an atom in +6 oxidation state, let's analyze each option given:

Option A: $MnO_4^-$

To determine the oxidation state of Mn in $MnO_4^-$, recall that the oxidation state of oxygen is typically -2. In the permanganate ion, there are four oxygens, contributing a total of $4 \times (-2) = -8$ to the charge. The overall charge of the $MnO_4^-$ ion is -1, which means the oxidation state of Mn must balance the charge contributed by oxygen to achieve the overall -1 charge. Thus, if we let the oxidation state of Mn be x, we have:

$x + 4(-2) = -1$

Solving for x gives:

$x - 8 = -1 \Rightarrow x = +7$

Option B: $Cr(CN)_6^{3-}$

For the $Cr(CN)_6^{3-}$ ion, cyanide $(CN)^-$ is a ligand with a -1 charge. With six cyanide ions, the total charge contributed by the ligands is $6 \times (-1) = -6$. The overall charge of the complex is -3, which means the oxidation state of Cr must offset the ligands' charge to achieve the overall -3 charge. Letting the oxidation state of Cr be x, we can set up the equation:

$x + 6(-1) = -3$

Solving for x gives:

$x - 6 = -3 \Rightarrow x = +3$

Option C: $NiF_6^{2-}$

In $NiF_6^{2-}$, the fluoride ion $(F^-)$ has an oxidation state of -1. With six fluoride ions, the total charge from the fluoride ions is $6 \times (-1) = -6$. For the complex to have an overall charge of -2, the oxidation state of Ni (let's call it x) needs to satisfy the following equation:

$x + 6(-1) = -2$

Solving for x gives:

$x - 6 = -2 \Rightarrow x = +4$

Option D: $CrO_2Cl_2$

In chromium dioxide dichloride, we have two oxygen atoms and two chlorine atoms bonded to chromium. Oxygen has a typical oxidation state of -2, and chlorine has a typical oxidation state of -1. However, being in a molecule (not an ion), the total must sum to zero. Let the oxidation state of Cr be x; we can set up the equation as follows:

$x + 2(-2) + 2(-1) = 0$

This simplifies to:

$x - 4 - 2 = 0 \Rightarrow x = +6$

Conclusion: From the analysis above, the species with a chromium atom in the +6 oxidation state is found in Option D: $CrO_2Cl_2$.

Normality (N) is a measure of concentration equivalent to the gram equivalent weight per liter of solution. For acids, it's determined by the amount of H⁺ ions that one molecule of acid can furnish in its reaction. Phosphorous acid, $H_3PO_3$, is a diprotic acid, meaning each molecule can furnish two H⁺ ions for reactions, despite having three hydrogen atoms, because only two of the hydrogen atoms are bonded in such a way that they can be ionized and participate in a reaction as protons ($H^+$).

To calculate the normality of a solution, you use the formula:

$N = M \times n$

where:

• $N$ is the normality,
• $M$ is the molarity of the solution, and
• $n$ is the number of moles of replaceable ions furnished by one mole of solute (in this case, the $H^+$ ions from phosphorus acid).

Given that the molarity ($M$) of phosphorus acid is 0.3 M, and it can furnish 2 $H^+$ ions per molecule, we use $n = 2$:

$N = 0.3 \,M \times 2 = 0.6\,N$

Therefore, the normality of the 0.3 M phosphorus acid ($H_3PO_3$) solution is 0.6 N. So, the correct option is:

Option D: 0.6

To determine which option corresponds to the scenario where the equivalent weight of $MnSO_4$ is half its molecular weight upon conversion, we need to analyze each oxidation process and find out the change in oxidation state of manganese (Mn) in each case. The equivalent weight of a compound in a given reaction depends on the change in oxidation number per atom of the element in question. It is calculated as the molecular weight divided by the number of electrons lost or gained in the reaction.

The molecular weight of $MnSO_4$ (Manganese(II) sulfate) is primarily irrelevant for the specifics of calculating equivalent weight in this manner; what's crucial is the stoichiometry of the electron transfer involved in its reaction.

Let's consider the oxidation state of Mn in $MnSO_4$, which is +2.

Option A: $Mn_2O_3$

Mn's oxidation state in $Mn_2O_3$ is +3. Thus, going from +2 in $MnSO_4$ to +3 in $Mn_2O_3$ involves a loss (or gain) of 1 electron per manganese atom. Given that there are two Mn atoms in $Mn_2O_3$, this would not match the stated requirement directly without considering the specifics of the individual reaction stoichiometry.

Option B: $MnO_2$

The oxidation state of Mn in $MnO_2$ is +4. The change from +2 to +4 implies a gain or loss of 2 electrons.

Option C: $MnO_4^-$ (Permanganate)

In $MnO_4^-$, the oxidation state of Mn is +7. The change from +2 in $MnSO_4$ to +7 in $MnO_4^-$ involves a change of 5 electrons.

Option D: $MnO_4^{2-}$

This oxidation state is not typically encountered for manganese and may represent a mistaken formulation, as Mn typically shows +7 oxidation state in permanganate $(MnO_4^-)$, not a doubly charged anion.

Given these considerations, the question's conditions mention that the equivalent weight of $MnSO_4$ becomes half of its molecular weight when it converts to one of these products. This would imply a situation where the number of electrons transferred per formula unit of $MnSO_4$ is equivalent to 2 (since the molecular weight divided by 2 equals the equivalent weight when each mole of $MnSO_4$ involves the transfer of 2 moles of electrons).

Thus, the correct answer is:

Option B: $MnO_2$, because the conversion from $Mn^{+2}$ to $Mn^{+4}$ involves a transfer of 2 electrons per manganese atom, which fits the condition that the equivalent weight is half the molecular weight in this context.