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23 October, 19:45

Consider the following reaction and select the false statement below.

10 I - (aq) + 16 H + (aq) + 2 MnO4 - (aq) → 5 I2 (s) + 2 Mn2 + (aq) + 8 H2O (l)

(1) this is a reduction-oxidation (redox) reaction

(2) the iodine ion is the reducing agent

(3) the permanganate ion is the oxidizing agent

(4) the hydrogen ion is neither reduced nor oxidized

(5) in the reaction as written, ten moles of electrons are transferred from the oxidizing agent to the reducing agent

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  1. 23 October, 21:07
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    10 I⁻ (aq) + 16 H⁺ (aq) + 2 MnO₄ - (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H₂O (l)

    Option 5 is the FALSE

    Explanation:

    Option 1: TRUE

    This a redox reaction, because one element is been reduced and another one is been oxidated. You must see the oxidation numbers of each element, to find out which one is the oxidizing and the reducing agent.

    Option 2: TRUE

    Iodide has been oxidated, so it's the reducing agent

    2I⁻ → I₂ + 2e⁻

    When the oxidation number increase, the element is been oxidated.

    Iodine changes from - 1 to 0

    Option 3: True

    Mn in permanganate has been reduced, so it's the oxidizing agent

    8H⁺ + MnO₄ - + 5e⁻ → 2Mn²⁺ + 4H₂O

    Mn change the oxidation number from 7 + to 2+, it has released 5e-

    Option 4: True

    Hydrogen ion stays the same, in both sides of the reaction (H⁺)

    Option 5: FALSE

    As we have 2 moles of e⁻ in half reaction of oxidation and 5e⁻, in reduction we have to multiply x5 and x2 each, so finally we get 10 moles of e⁻.

    But this 10e⁻ are transferred from the reducing agent (which is the one that release them) to the oxidizing agent (which is the one that catch them, to decrease the oxidation number)
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