Question

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

Answer 1

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

Option 5 is the FALSE

Step-by-step 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)