Redox Reactions

The reaction is

${\mathrm{MnO}}_4^{-}$ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

Using N₁V₁ = N₂V₂,

$\frac{1}{8}\times 5V_1=\frac{1}{4}\times 25$

Solving, V₁ = 10 mL.

Balanced redox equation in alkaline medium:

Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O

To balance:

Unbalanced form given in question:

aCl₂ + bOH⁻ → cClO⁻ + dCl⁻ + eH₂O

Balanced form:

a = 1, b = 2, c = 1, d = 1, e = 1

Hence, the correct option is: A (1, 2, 1, 1).

$A_{2}B\to 2A^{+}+B^{-2},B^{-2}$ has complete octet in its dianionic form, thus in its atomic state it has 6 electrons in its valence shell. As it has negative charge, it has acquired two electrons to complete its octet.

• (a) $\mathrm{Fe^{3+}}$ oxidises $\mathrm{I^-}$ to $\mathrm{I_2}$ and itself gets reduced to $\mathrm{Fe^{2+}}$ → correct redox reaction.

• (b) $\mathrm{FeI_3}$ is not stable in aqueous solution → incorrect.

• (c) $\mathrm{Fe^{2+}}$ cannot be reduced to Fe (0) by $\mathrm{S_2O_8^{2-}}$ → incorrect.

• (d) $\mathrm{S_2O_8^{2-}}$​ is a strong oxidant and oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$ → correct.

Balanced redox equation in alkaline medium:

Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O

To balance:

Unbalanced form given in question:

aCl₂ + bOH⁻ → cClO⁻ + dCl⁻ + eH₂O

Balanced form:

a = 1, b = 2, c = 1, d = 1, e = 1

Hence, the correct option is: A (1, 2, 1, 1).

Given half-reaction in acidic medium:

Cr₂O₇²⁻ + XH⁺ + Ye⁻ → 2A + ZH₂O

We know standard reduction of dichromate in acidic medium:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Now comparing,

X = number of H⁺ = 14
Y = number of electrons = 6
Z = number of H₂O = 7
A = product formed = Cr³⁺

So values are:
X = 14, Y = 6, Z = 7, A = Cr³⁺

Given half-reaction in acidic medium:

Cr₂O₇²⁻ + XH⁺ + Ye⁻ → 2A + ZH₂O

We know standard reduction of dichromate in acidic medium:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Now comparing,

X = number of H⁺ = 14
Y = number of electrons = 6
Z = number of H₂O = 7
A = product formed = Cr³⁺

So values are:
X = 14, Y = 6, Z = 7, A = Cr³⁺

• (a) $\mathrm{Fe^{3+}}$ oxidises $\mathrm{I^-}$ to $\mathrm{I_2}$ and itself gets reduced to $\mathrm{Fe^{2+}}$ → correct redox reaction.

• (b) $\mathrm{FeI_3}$ is not stable in aqueous solution → incorrect.

• (c) $\mathrm{Fe^{2+}}$ cannot be reduced to Fe (0) by $\mathrm{S_2O_8^{2-}}$ → incorrect.

• (d) $\mathrm{S_2O_8^{2-}}$​ is a strong oxidant and oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$ → correct.

In ${\mathrm{N}}^{3-}$ ion 'N' is present in its lowest possible oxidation state, hence it cannot be reduced further because of which it cannot act as an oxidizing agent.

In ${\mathrm{N}}^{3-}$ion 'N' is present in its lowest possible oxidation state, hence it cannot be reduced further because of which it cannot act as an oxidizing agent.

In ${\mathrm{N}}^{3-}$ion 'N' is present in its lowest possible oxidation state, hence it cannot be reduced further because of which it cannot act as an oxidizing agent.

Correct explanation: Bromine is a stronger oxidising agent than iodine, isliye Br₂ thiosulphate ko oxidise karta hai jabki I₂ usse reduce hota hai. Isliye thiosulphate ka behaviour Br₂ aur I₂ ke saath alag hota hai.

Correct option: C – Bromine is a stronger oxidant than iodine.

$S{O}_3\Rightarrow S+3\times (-2)=0$

$\boxed{S=+6}$

$H_{2}S{O}_3\Rightarrow 2\times (+1)+S+3\times (-2)=0$

$\boxed{S=+4}$

$SOC{l}_2\Rightarrow S+1\times (-2)+2\times (-1)=0$

$\boxed{S=+4}$

$S{F}_4\Rightarrow S+4\times (-1)=0$

$\boxed{S=+4}$

$BaS{O}_4\Rightarrow +2+S+4\times (-2)=0$

$\boxed{S=+6}$

$H_2{S}_2{O}_7\Rightarrow 2\times (+1)+2S+7\times (-2)=0$

$\boxed{S=+6}$

The reaction is

${\mathrm{MnO}}_4^{-}$ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

Using N₁V₁ = N₂V₂,

$\frac{1}{8}\times 5V_1=\frac{1}{4}\times 25$

Solving, V₁ = 10 mL.

Oxidising agent wahi ban sakta hai jo khud reduce ho sake. N³⁻ me nitrogen ki oxidation state −3 hoti hai jo minimum hoti hai, aur ye aur reduce nahi ho sakta. Baaki ions (SO₄²⁻, BrO₃⁻, MnO₄⁻) me central atom ki high oxidation state hoti hai jo reduce ho sakti hai, isliye wo oxidising agent ho sakte hain.

Correct answer: N³⁻ cannot function as oxidising agent.

Oxidising agent wahi ban sakta hai jo khud reduce ho sake. N³⁻ me nitrogen ki oxidation state −3 hoti hai jo minimum hoti hai, aur ye aur reduce nahi ho sakta. Baaki ions (SO₄²⁻, BrO₃⁻, MnO₄⁻) me central atom ki high oxidation state hoti hai jo reduce ho sakti hai, isliye wo oxidising agent ho sakte hain.

Correct answer: N³⁻ cannot function as oxidising agent.

In the oxalic acid $\left(H_2{C}_2{O}_4\right)$ vs $KMn{O}_4$ titration, the oxidation of oxalate $\left(C_2{O_4}^{2-}\right)$ to $C{O}_2$ is slow at room temperature.

$5C_2{O}_4^{2-}+2Mn{O}_4^{-}+16H^{+}\to 10C{O}_2+$
$2M{n}^{2+}+8H_{2}O$
Why heating is needed?
- The initial formation of $M{n}^{2+}$ from $KMn{O}_4$ is slow at room temperature.
- $M{n}^{2+}$ formed acts as a catalyst, accelerating the rest of the reaction.
- Hence, the solution is heated to around $60°C$ to speed up the reaction.
In $FAS\left(F{e}^{2+}\right)$ vs $KMn{O}_4$ titration, the reaction proceeds quickly at room temperature:

$5F{e}^{2+}+Mn{O}_4^{-}+8H^{+}$$\to 5F{e}^{3+}+M{n}^{2+}+4H_{2}O$

Why no heating is needed?
- The oxidation of $F{e}^{2+}$ is already fast at room temperature.
- Heating can cause $F{e}^{2+}$ to be oxidized to $F{e}^{3+}$ by air, leading to errors.
Hence, both statements are true.

${{\mathrm{ClO}}_4}^{-}$ does not undergo disproportionation reaction because chlorine is present in it's highest oxidation state i.e. +7.

${{\mathrm{ClO}}_4}^{-}$ → chlorine is in its highest oxidation state i.e., +7.
Chlorine can exhibit a –1 to +7 oxidation state.
The oxidation states of chlorine which can undergo disproportionation are : 0, +1, +3, and +5.

Correct explanation: Bromine is a stronger oxidising agent than iodine, isliye Br₂ thiosulphate ko oxidise karta hai jabki I₂ usse reduce hota hai. Isliye thiosulphate ka behaviour Br₂ aur I₂ ke saath alag hota hai.

Correct option: C – Bromine is a stronger oxidant than iodine.

In the oxalic acid $\left(H_2{C}_2{O}_4\right)$ vs $KMn{O}_4$ titration, the oxidation of oxalate $\left(C_2{O_4}^{2-}\right)$ to $C{O}_2$ is slow at room temperature.

$5C_2{O}_4^{2-}+2Mn{O}_4^{-}+16H^{+}\to 10C{O}_2+$
$2M{n}^{2+}+8H_{2}O$
Why heating is needed?
- The initial formation of $M{n}^{2+}$ from $KMn{O}_4$ is slow at room temperature.
- $M{n}^{2+}$ formed acts as a catalyst, accelerating the rest of the reaction.
- Hence, the solution is heated to around $60°C$ to speed up the reaction.
In $FAS\left(F{e}^{2+}\right)$ vs $KMn{O}_4$ titration, the reaction proceeds quickly at room temperature:

$5F{e}^{2+}+Mn{O}_4^{-}+8H^{+}$$\to 5F{e}^{3+}+M{n}^{2+}+4H_{2}O$

Why no heating is needed?
- The oxidation of $F{e}^{2+}$ is already fast at room temperature.
- Heating can cause $F{e}^{2+}$ to be oxidized to $F{e}^{3+}$ by air, leading to errors.
Hence, both statements are true.

BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl is a double displacement reaction with no change in oxidation states, so it is not a redox reaction. In the other options, there is a clear change in oxidation numbers indicating redox processes.

In the oxalic acid $\left(H_2{C}_2{O}_4\right)$ vs $KMn{O}_4$ titration, the oxidation of oxalate $\left(C_2{O_4}^{2-}\right)$ to $C{O}_2$ is slow at room temperature.

$5C_2{O}_4^{2-}+2Mn{O}_4^{-}+16H^{+}\to 10C{O}_2+$
$2M{n}^{2+}+8H_{2}O$
Why heating is needed?
- The initial formation of $M{n}^{2+}$ from $KMn{O}_4$ is slow at room temperature.
- $M{n}^{2+}$ formed acts as a catalyst, accelerating the rest of the reaction.
- Hence, the solution is heated to around $60°C$ to speed up the reaction.
In $FAS\left(F{e}^{2+}\right)$ vs $KMn{O}_4$ titration, the reaction proceeds quickly at room temperature:

$5F{e}^{2+}+Mn{O}_4^{-}+8H^{+}\to 5F{e}^{3+}+M{n}^{2+}+4H_{2}O$

Why no heating is needed?
- The oxidation of $F{e}^{2+}$ is already fast at room temperature.
- Heating can cause $F{e}^{2+}$ to be oxidized to $F{e}^{3+}$ by air, leading to errors.
Hence, both statements are true.

${{\mathrm{ClO}}_4}^{-}$ does not undergo disproportionation reaction because chlorine is present in it's highest oxidation state i.e. +7.

Correct explanation: Bromine is a stronger oxidising agent than iodine, isliye Br₂ thiosulphate ko oxidise karta hai jabki I₂ usse reduce hota hai. Isliye thiosulphate ka behaviour Br₂ aur I₂ ke saath alag hota hai.

Correct option: C – Bromine is a stronger oxidant than iodine.

(2)

$3{\mathrm{Cl}}_2+6{\mathrm{OH}}^{-}\to 5{\mathrm{Cl}}^{-}+{\mathrm{ClO}}_3^{-}+3{\mathrm{H}}_2\mathrm{O}$

In the oxalic acid $\left(H_2{C}_2{O}_4\right)$ vs $KMn{O}_4$ titration, the oxidation of oxalate $\left(C_2{O_4}^{2-}\right)$ to $C{O}_2$ is slow at room temperature.

$5C_2{O}_4^{2-}+2Mn{O}_4^{-}+16H^{+}\to 10C{O}_2+$
$2M{n}^{2+}+8H_{2}O$
Why heating is needed?
- The initial formation of $M{n}^{2+}$ from $KMn{O}_4$ is slow at room temperature.
- $M{n}^{2+}$ formed acts as a catalyst, accelerating the rest of the reaction.
- Hence, the solution is heated to around $60°C$ to speed up the reaction.
In $FAS\left(F{e}^{2+}\right)$ vs $KMn{O}_4$ titration, the reaction proceeds quickly at room temperature:

$5F{e}^{2+}+Mn{O}_4^{-}+8H^{+}\to 5F{e}^{3+}+M{n}^{2+}+4H_{2}O$

Why no heating is needed?
- The oxidation of $F{e}^{2+}$ is already fast at room temperature.
- Heating can cause $F{e}^{2+}$ to be oxidized to $F{e}^{3+}$ by air, leading to errors.
Hence, both statements are true.