Redox Reactions

First, separate into two half-reactions:

Oxidation $\left(\mathrm{Fe}^{2+}\right.$ to $\left.\mathrm{Fe}^{3+}\right)$,

$ \mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+1 e^{-} $

Reduction ( $\mathrm{MnO}_4^{-}$to $\mathrm{Mn}^{2+}$ ),

$ \mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+} $

Balance O by adding $4 \mathrm{H}_2 \mathrm{O}$,

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

Balance H by adding $8 \mathrm{H}^{+}$,

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

Add $5 e^{-}$to balance charge,

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

Now, multiply oxidation half-reaction by 5 :

$ 5 \mathrm{Fe}^{2+} \longrightarrow 5 \mathrm{Fe}^{3+}+5 e^{-} $

Now add both:

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

Thus, the stoichiometric coefficient of $\mathrm{H}_2 \mathrm{O}=4$

Balanced reaction,

$ \begin{aligned} & 5 \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+} \longrightarrow 5 \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\ & \text { Using, molarity }=\frac{\text { Number of moles of solute }}{\text { Volume of solution }(\mathrm{L})} \\ & \begin{aligned} \text { or } \quad\left(\mathrm{KMnO}_4\right) & =\left(\mathrm{FeSO}_4\right) \\ \frac{M_1 V_1}{n_1} & =\frac{M_2 V_2}{n_2} \\ \mathrm{Fe}^{2+} \text { moles } & =0.02 \times 0.5=0.01 \mathrm{~mol} \end{aligned} \end{aligned} $

5 moles of $\mathrm{Fe}^{2+}$ react with 1 mole of $\mathrm{KMnO}_4$. So,

$\mathrm{KMnO}_4$ required $=0.01 / 5=0.002 \mathrm{~mol}$

Volume $=0.002 / 0.1=0.02 \mathrm{~L}=20 \mathrm{~mL}$

Oxidation half reaction:

$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HSO}_{4}^{-}(a q)+3 \mathrm{H}^{+}(a q)+2 e^{-} \quad \ldots \text { (i) } $

Reduction half reaction:

$ \mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \quad \ldots \text { (ii) } $

Multiply eq. (i) by 5 and (ii) by 2 and then add

$ \begin{aligned} 2 \mathrm{MnO}_{4}^{-}(a q)+5 \mathrm{SO}_{2}(g) & +2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Mn}^{+}(a q)+5 \mathrm{HSO}_{4}^{-}(a q) \end{aligned} $

Thus, the stoichiometric coefficient of $ \mathrm{SO}_{2} $ in the above balance equation is 5 .

To determine the volume of $ \frac{M}{8} \mathrm{KMnO_4} $ solution needed to completely react with $ 25.0 \, \mathrm{cm}^3 $ of $ \frac{M}{4} \mathrm{FeSO_4} $ in an acidic medium, we start by considering the balanced ionic equation:

$ \mathrm{MnO_4}^- + 5 \mathrm{Fe}^{2+} + 8 \mathrm{H}^+ \rightarrow \mathrm{Mn}^{2+} + 5 \mathrm{Fe}^{3+} + 4 \mathrm{H_2O} $

From this equation, it's evident that 1 mole of $ \mathrm{KMnO_4} $ reacts with 5 moles of $ \mathrm{FeSO_4} $.

For balancing the redox reaction, apply the molarity equation:

$ \frac{M_1 V_1}{n_1}(\mathrm{KMnO_4}) = \frac{M_2 V_2}{n_2}(\mathrm{FeSO_4}) $

Substitute the given values where:

$ M_1 = \frac{M}{8} $ (molarity of $\mathrm{KMnO_4}$)

$ M_2 = \frac{M}{4} $ (molarity of $\mathrm{FeSO_4}$)

$ V_2 = 25 \, \mathrm{cm}^3 $

$ n_1 = 1 $ (moles of $\mathrm{KMnO_4}$)

$ n_2 = 5 $ (moles of $\mathrm{FeSO_4}$)

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

Solving the equation for $ V_1 $:

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

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

$ V_1 = 10 \, \mathrm{cm}^3 \quad \text{or} \quad V_1 = 10.0 \, \mathrm{mL} $

Hence, the volume of $ \frac{M}{8} \mathrm{KMnO_4} $ solution required is 10.0 mL.

Perhydrol means 30% solution of H₂O₂.

H₂O₂ decomposes as :

2H₂O₂ $\to$ 2H₂O + O₂

Volume strength of 30% H₂O₂ solution is 100 that means 1 mL of this solution on decomposition given 100 mL oxygen.

Since, 100 mL of oxygen is obtained by = 1 mL of H₂O₂

$\therefore$ 1000 mL of oxygen will be obtained by

$ = {1 \over {100}} \times 1000$ mL of H₂O₂

= 10 mL of H₂O₂

Chemical reaction given in option 'c' is correct. Rest of the given chemical reactions are incorrect. These can be corrected as:

2NH₃ + CaSO₄ + CO₂ + H₂O $\to$ (NH₄)₂SO₄ + CaCO₃

CaC₂ + H₂O $\to$ 2C₂H₂ + Ca(OH)₂

Na₂S₂O₃ + H₂SO₄ $\to$ Na₂SO₄ + SO₂ + S₈ + H₂O

Oxidation number of S in Na₂S₂O₃ = + 2

Oxidation number of S in H₂SO₄ = + 6

Oxidation number of S in SO₂ = + 4

Oxidation number of S in Na₂SO₄ = + 6

Oxidation number of S in S₈ = 0

So, sulphur can oxidise from + 2 to + 4 as well as reduce to 0 (S₈).

5C₂O$_4^{2 - }$ + 2MnO$_4^{ - }$ + 16H⁺ $\to$ 2Mn²⁺ + 10CO₂ + 8H₂O

Note In a reaction (redox/neutralisation) number of equivalents of different reactants are same.

Equivalent weight of C₂O$_4^{2 - }$ = ${{Molecular\,weight} \over 2}$

$\therefore$ Molarity = 2 $\times$ Normality

Equivalent weight of MnO$_4^{ - }$ = ${{Molecular\,weight} \over 5}$

Molarity = 5 $\times$ Normality

Number of equivalents of MnO$_4^{- }$ = ${{28.85 \times 5\,{M_1}} \over {1000}}$

Number of equivalents of Na₂C₂O$_4^{- }$ = ${{0.1467} \over {67}}$

$\therefore$ ${{28.85 \times 5\,{M_1}} \over {1000}} - {{0.1467} \over {67}}$

M₁ = 0.01518 M