Question

Suppose 20 ml of a 0.15 M C₂O₄²⁻ solution is reacted with an excess amount of permanganate ion. How many electrons (in moles) are produced during the course of this reaction?

Answer 1

In the given reaction, 0.15 M C₂O₄²⁻ solution is reacted with an excess amount of permanganate ion. The reaction produces 0.015 moles of electrons.

Step-by-step explanation:

In the given reaction, the permanganate ion (MnO4-) is reduced to manganese (IV) oxide (MnO2), and during this reduction, the permanganate ion gains 5 electrons.

1 mole of electrons is equal to 6.022 × 1023 electrons.

Therefore, the number of moles of electrons produced during the reaction can be calculated as:

(0.15 M C2O42-) × [(20 ml) / (1000 ml/1 L)] × [(5 mol electrons) / (1 mol C2O42-)] = 0.015 moles of electrons.