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What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the overall redox equation: 5 ag+(aq) + mn2+(aq) + 4h2o(l) → 5ag(s) + mno4–(aq) + 8h+(aq)?

Respuesta :

transitionstate
The equation is as below,

       5 Ag⁺ (aq) + Mn²⁺ (aq) + 4H₂O (l) → 5Ag (s) + MnO4⁻(aq) + 8H⁺ (aq)

Reduction:
                According to eq. Silver is reduced as,

                                5 Ag⁺  + 5 e⁻    →    5 Ag

One electron is gained per Ag ion, so 5 Ag ions will gain 5 electrons.

Oxidation:
               According to eq. Manganese is Oxidized as, 

                                        Mn²⁺    →    MnO4⁻

Oxidation number of Mn in MnO₄⁻ is +7, so Mn²⁺ is loosing 5 electrons, Hence,

                                        Mn²⁺    →    MnO4⁻  + 5 e⁻

Results:
             Hence, 5 moles of electrons are exchanged between oxidizing and reducing agents.
mickymike92

Based on the equation of the reaction, 5 moles of electrons are exchanged between oxidizing and reducing agents.

What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent?

The equation of the equation is as follows:

  • 5 Ag⁺ (aq) + Mn²⁺ (aq) + 4H₂O (l) → 5Ag (s) + MnO4⁻(aq) + 8H⁺ (aq)

The reduction equation is as follows:

5 Ag⁺  + 5 e⁻    →    5 Ag

5 moles of electrons are accepted by silver ions

The oxidation equation is as follows:

Mn²⁺    →    MnO4⁻

Mn²⁺ is oxidized to MnO₄⁻ is +7 by loosing 5 electrons.

Therefore, 5 moles of electrons are exchanged between oxidizing and reducing agents.

Learn mores about reduction and oxidation at: https://brainly.com/question/4222605

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