9 Ethanedioic acid, HO₂CCO₂H, can be oxidised by KMnO4 in dilute sulfuric acid. The products of this reaction are carbon dioxide, water, potassium sulfate and manganese(II) sulfate. In this reaction each ethanedioic acid molecule loses two electrons as it is oxidised. A half-equation for this process is shown. HO₂CCO₂H →→ 2CO₂ + 2H+ + 2e™ How many water molecules are produced when five ethanedioic acid molecules are oxidised by KMnO4 in dilute sulfuric acid? A 5 B 8 C 10 D 16​

Respuesta :

The oxidation reaction is the chemical reaction in which the species loses electrons. The five oxidized ethanedioic acid molecules will produce 8 water molecules. Thus, option B is correct.

What is the half-reaction equation?

Half reaction equations are the representation of the reduction and the oxidation half of the whole reaction that denotes the gaining and losing of electrons differently.

The oxidation half of the reaction is given as:

HO₂CCO₂H → 2CO₂ + 2H⁺ + 2e⁻

The reduction half of the reaction will be given as:

MnO₄ + H⁺ → Mn²⁺ + H₂O

The overall reaction is written as:

C₂O₄ ⁻² + MnO₄⁻ + H⁺ →  CO₂ + H₂O + Mn²⁺

Here electrons are lost in the formation of carbon dioxide and electrons are gained when manganese is produced.

The total reaction is given as:

5 C₂O₄ ²⁻ + 2MnO₄ ⁻ + 16H⁺ → 10CO₂ + 8 H₂O + 2 Mn²⁺

Therefore, option B. five ethanedioic acid molecules produce 8 water molecules.

Learn more about redox half-reactions here:

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