Question

Observe the following reaction: \[ xI^- + y \, \, MnO_4^- + z \, H^+ \rightarrow a \, Mn^{2+} + b \, H_2O + cl_2 \] Which of the following are correct? (i) \( y : x = 2 : 5 \)
(ii) \( y : a = 1 : 1 \)
(iii) \( x : c = 1 : 2 \)
(iv) \( y : e = 2 : 5 \)

• A. i, ii, iii only
• B. ii, iii, iv only
• C. i, ii, iv only
• D. i, iii, iv only

Hint:

Balancing redox reactions requires ensuring that the number of atoms and charges are equal on both sides of the equation.

Answer 1

The Correct Option is C

Step 1: Assigning Oxidation States - Iodide ion \( I^- \) is oxidized to iodine \( I_2 \). - Manganate ion \( MnO_4^- \) is reduced to \( Mn^{2+} \). - \( H^+ \) ions balance the charge and form water (\( H_2O \)). 

Step 2: Writing the Half-Reactions Oxidation Half-Reaction (Iodine Ion): \[ 2I^- \rightarrow I_2 + 2e^- \] Reduction Half-Reaction (Permanganate Ion): \[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \] 

Step 3: Balancing Electrons - Multiply the oxidation reaction by 5 and the reduction reaction by 2 to balance the electrons: \[ 10I^- \rightarrow 5I_2 + 10e^- \] \[ 2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O \] 

Step 4: Writing the Overall Balanced Equation \[ 10I^- + 2MnO_4^- + 16H^+ \rightarrow 5I_2 + 2Mn^{2+} + 8H_2O \] 

Step 5: Verifying the Given Ratios - (i) \( y : x = 2 : 5 \) - \( y = 2 \) (MnO\(_4^-\)) and \( x = 10 \) (I\(^-\)), so the ratio simplifies to 2:5. - (ii) \( y : a = 1 : 1 \) - \( y = 2 \) and \( a = 2 \), so the ratio is 1:1. - (iii) \( x : c = 1 : 2 \) - Given equation shows \( x = 10 \) and \( c = 5 \), so the ratio is 10:5 = 2:1, not 1:2. - (iv) \( y : e = 2 : 5 \) - \( y = 2 \) (MnO\(_4^-\)) and \( e = 5 \) (I\(_2\)), so the ratio is 2:5. 

Step 6: Conclusion Since (i), (ii), and (iv) are correct, the correct answer is option (3).