Redox Reactions

$\therefore$ Equivalent of KMnO4 = Eq of FeC2O4 + Eq of Fe2(C2O4)3 + FeSO4 $\Rightarrow$ moles of KMnO4 $\times$ 5 = 3 $\times$ 1 + 6 $\times$ 1 + 1 $\times$ 1 $\Rightarrow$ moles of KMnO4 =

= 2

In the given reactions:

In the first reaction with iodine ($I_2$), thiosulfate ($S_2O_3^{2-}$) acts as a reducing agent, converting $I_2$ to $I^{-}$, while it is itself oxidized to $S_4O_6^{2-}$.

This indicates that iodine is acting as an oxidant (oxidizing agent), getting reduced in the process.

In the second reaction with bromine ($Br_2$), thiosulfate is oxidized more extensively, being converted all the way to sulfate ($SO_4^{2-}$), with hydrogen ions ($H^+$) and bromide ions ($Br^-$) also produced.

This shows that bromine acts as a stronger oxidant compared to iodine, as it promotes a complete oxidation of $S_2O_3^{2-}$ to $SO_4^{2-}$.

Given these observations, the correct option that justifies the dual behavior of thiosulfate reacting with iodine and bromine differently is: Option C: Bromine is a stronger oxidant than iodine This statement correctly explains why thiosulfate is oxidized to $S_4O_6^{2-}$ in the presence of iodine, and to $SO_4^{2-}$ in the presence of bromine.

Bromine's stronger oxidizing ability leads to a more complete oxidation of thiosulfate compared to the oxidation by iodine.

(A) Combustion of hydrocarbon (B) Decomposition into gaseous product.

(C) Displacement of ' $V$ ' by ' Ca ' atom.

(D) Disproportionation of $\mathrm{H}_2 \mathrm{O}_2^{-1}$ into $\mathrm{O}^{-2}$ and $\mathrm{O}^{\circ}$ oxidation states.

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

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 :

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 $\text{H}_2\text{C}_2\text{O}_4$ :

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

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 :

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

NaH is a strong H– (hydride) donor.

Hence cannot be used as an oxidising agent.

In Pyridine, the lone pair of ‘N’ is localised, makes it basic.

When a salt reacts with

and concentrated

, the type of gas evolved depends on the anion present in the salt.

In this case, a greenish yellow gas is evolved, which indicates the formation of chlorine gas (

).

gas is greenish-yellow in color and is typically produced from halide salts (namely chlorides, bromides, or iodides) when they are oxidized.

Among the given options, we need to identify a salt that contains a halide which can produce

upon reaction with

and concentrated

. Option A (

) contains chloride ions, and when heated with

and concentrated

, it can undergo a redox reaction where the chloride ions are oxidized to

gas. The corresponding reaction can be represented as follows:

Option B (

) contains iodide ions, which would lead to the liberation of iodine or

, not a greenish yellow gas. Option C (

) contains nitrate ions and does not produce a halogen gas upon reaction with

and

. Option D (

) contains bromide ions which would lead to the formation of bromine (

), a reddish-brown gas, not greenish yellow. Therefore, the correct answer is Option A (

), since it is the salt that can produce a greenish yellow gas (

) when reacted with

and concentrated

.

A disproportionation reaction is a chemical reaction in which a single substance is simultaneously reduced and oxidized, forming two different products.

To identify disproportionation reactions, one should look for a specified element in the reactant that gains electrons (reduction) and for the same element that lose electrons (oxidation).

Here's an analysis of each reaction : (A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$ Here, copper (Cu) is in the +1 oxidation state and forms two products, where it is oxidized to the +2 state in $\mathrm{Cu}^{2+}$ and reduced to the 0 state in Cu.

This is a classic example of disproportionation.

So, (A) is true for disproportionation.

(B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$ In this reaction, $\mathrm{MnO}_4^{2-}$ (where Mn is in +6 oxidation state) is converted into $\mathrm{MnO}_4^{-}$ (where Mn is in +7 oxidation state, oxidation has occurred) and $\mathrm{MnO}_2$ (where Mn is in +4 oxidation state, reduction has occurred).

Here, again, we have the same element, manganese (Mn), undergoing both reduction and oxidation.

Hence, (B) is also a disproportionation reaction.

(C) $2 \mathrm{KMnO}_4 \longrightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2$ For the reaction with potassium permanganate, $\mathrm{KMnO}_4$, manganese starts in the +7 oxidation state.

It forms $\mathrm{K}_2 \mathrm{MnO}_4$ where Mn is in +6 state (reduction) and $\mathrm{MnO}_2$ where Mn is in +4 state (further reduction), but no increase in oxidation state of manganese is observed.

Instead, oxygen is released, so the manganese is not being oxidized in any of its products; it is only reduced twice.

This is not a disproportionation reaction because the same element should be undergoing both oxidation and reduction, which is not the case here.

Hence, (C) is not a disproportionation reaction.

(D) $2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}$ In this reaction, the $\mathrm{MnO}_4^{-}$ ion (where Mn is at +7 oxidation state) and $\mathrm{Mn}^{2+}$ ion (where Mn is at +2 oxidation state) react to form $\mathrm{MnO}_2$ (where Mn is at +4 oxidation state).

This reaction is not a case of disproportionation; it is a comproportionation or synproportionation reaction, where two species with the same element in different oxidation states produce a product with the element at an intermediate oxidation state.

Therefore, the disproportionation reactions among the given choices are: (A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$ (B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$ Option A would be the correct choice: (A), (B).

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The reducing agent loses electron during redox reaction i.e. oxidises itself.