Question:

Consider the following equations :
\(\mathrm{2Fe^{2+} + H_2O_2 \to xA + yB}\) (in basic medium)
\(\mathrm{2MnO_4^- + 6H^+ + 5H_2O_2 \to x'C + y'D + z'E}\) (in acidic medium)
The sum of the stoichiometric coefficients x, y, x', y', and z' for products A, B, C, D and E, respectively, is ______.

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Fe2+ goes to Fe3+ (2 electrons lost for 2 Fe) while \(\mathrm{H_2O_2}\) gains 2 electrons to give 2OH-; in acid, \(\mathrm{MnO_4^-}\) goes to \(\mathrm{Mn^{2+}}\) and \(\mathrm{H_2O_2}\) is oxidised to \(\mathrm{O_2}\).
Updated On: Nov 5, 2026
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Correct Answer: 19

Solution and Explanation

Basic medium: \(\mathrm{2Fe^{2+} + H_2O_2 \to 2Fe^{3+} + 2OH^-}\). Check: charge left +4, right +6 - 2 = +4; two Fe lose 2 e-, \(\mathrm{H_2O_2}\) gains 2 e-. So A = Fe3+ (x = 2), B = OH- (y = 2).
Acidic medium: \(\mathrm{2MnO_4^- + 6H^+ + 5H_2O_2 \to 2Mn^{2+} + 8H_2O + 5O_2}\). Check: Mn 2 = 2; O: 8 + 10 = 18 on the left and 8 + 10 = 18 on the right; H: 6 + 10 = 16 = 8 x 2; charge: -2 + 6 = +4 = 2 x 2. So C = Mn2+ (x' = 2), D = H2O (y' = 8), E = O2 (z' = 5).
Sum = 2 + 2 + 2 + 8 + 5 = 19.
Answer: 19
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