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Introduction

An endless nu...

Question

An endless number of balanced equations can be written for the reaction between the permanganate ion and hydrogen peroxide in acidic solution to form the manganese (II) ion and oxygen:

$M n O_{4}^{-} (a q) + H_{2} O_{2} (a q) {H - \to}^{+} M n^{2 +} (a q) + O_{2} (g)$

Use the half-reaction method to determine the correct stoichiometry for this reaction.

2 M n O_{4}^{-} (a q) + 5 H_{2} O_{2} (a q) + 6 H^{+} (a q) \to 2 M n^{2 +} (a q) + 5 O_{2} (g) + 8 H_{2} O (l)

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3 M n O_{4}^{-} (a q) + 2 H_{2} O_{2} (a q) + 6 H^{+} (a q) \to 2 M n^{2 +} (a q) + 3 O_{2} (g) + 8 H_{2} O (l)

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4 M n O_{4}^{-} (a q) + 2 H_{2} O_{2} (a q) + 6 H^{+} (a q) \to 2 M n^{2 +} (a q) + 4 O_{2} (g) + 7 H_{2} O (l)

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None of the above

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Solution

The correct option is A $2 M n O_{4}^{-} (a q) + 5 H_{2} O_{2} (a q) + 6 H^{+} (a q) \to 2 M n^{2 +} (a q) + 5 O_{2} (g) + 8 H_{2} O (l)$
$M n O_{4}^{-} (a q) + H_{2} O_{2} (a q) {H - \to}^{+} M n^{2 +} (a q) + O_{2} (g)$

(1) First, divide the equation into two halves; an oxidation half-reaction and reduction half- reaction by grouping appropriate species.
$M n O_{4}^{-} \to M n^{2 +}$
$H_{2} O_{2} \to O_{2}$

(2) If necessary, balance both equations by inspection. In doing this ignore any oxygen and hydrogen atoms in the formula units. In other words, balance the non-hydrogen and non-oxygen atoms only.

(3) The third step involves balancing oxygen atoms. To do this, one must use water $(H_{2} O)$ molecules. Use 1 molecule of water for each oxygen atom that needs to be balanced. Add the appropriate number of water molecules to that side of the equation required to balance the oxygen atoms as shown below.
$M n O_{4}^{-} \to M n^{2 +} + 4 H_{2} O$
$H_{2} O_{2} \to O_{2}$

(4) Balancing the hydrogen atoms.
$M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O$
$H_{2} O_{2} \to O_{2} + 2 H^{+}$

(5) The fifth step involves the balancing of positive and negative charges.
$[5 e^{-} + M n O_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O] \times 2$
$[H_{2} O_{2} \to O_{2} + 2 H^{+} + 2 e^{-}] \times 5$

Now add both the reactions-
$2 M n O_{4}^{-} (a q) + 5 H_{2} O_{2} (a q) + 6 H^{+} (a q) \to 2 M n^{2 +} (a q) + 5 O_{2} (g) + 8 H_{2} O (l)$

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Similar questions

E^{o}

10 C l^{-} (a q) + 2 M n O_{4}^{-} (a q) + 16 H^{+} (a q) \to 5 C l_{2} (g) + 2 M n^{2 +} (a q) + 8 H_{2} O (l)

0.15 V

_{c}

10 C l^{-} (a q) + 2 M n O_{4}^{-} (a q) + 16 H^{+} (a q) \to 5 C l_{2} (g) + 2 M n^{2 +} (a q) + 8 H_{2} O (l)

E^{o}

25^{o}

2 M n^{2 +} (a q) + 4 O H^{-} (a q) + O_{2} (g) \to 2 M n O_{2} (s) + 2 H_{2} O (l)

M n O_{2} (s) + 2 I^{-} (a q) + 4 H^{+} (a q) \to M n^{2 +} (a q) + I_{2} (a q) + 2 H_{2} O (l)

2 S_{2} O_{3}^{2 -} (a q) + I_{2} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

2 M n^{2 +} (a q) + 4 O H^{-} (a q) + O_{2} (g) \to 2 M n O_{2} (s) + 2 H_{2} O (l)

M n O_{2} (s) + 2 I^{-} (a q) + 4 H^{+} (a q) \to M n^{2 +} (a q) + I_{2} (a q) + 2 H_{2} O (l)

2 S_{2} O_{3}^{2 -} (a q) + I_{2} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

S_{2} O_{3}^{2 -}

O_{2}

2 M n^{2 +} (a q) + 4 O H^{-} (a q) + O_{2} (g) \to 2 M n O_{2} (s) + 2 H_{2} O (l)

M n O_{2} (s) + 2 I^{-} (a q) + 4 H^{+} (a q) \to M n^{2 +} (a q) + I_{2} (a q) + 2 H_{2} O (l)

2 S_{2} O_{3}^{2 -} (a q) + I_{2} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

3 \times 10^{3}

O_{2}

I^{-}

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