In the electrosynthesis potassium manganate(VII) is converted to manga...
Explanation:
In electrosynthesis, the conversion of potassium manganate(VII) to manganese(IV) dioxide takes place. The process involves the passage of 1F of electrolysis.
Formula used:
1 mole of potassium manganate(VII) (KMnO4) on electrolysis gives 0.33 mole of manganese(IV) dioxide (MnO2).
Reasoning:
- 1F of electrolysis is the amount of electrical charge required to deposit one mole of an element during electrolysis.
- The formula used states that 1 mole of KMnO4 gives 0.33 mole of MnO2. This formula is derived from the balanced chemical equation for the reaction that takes place during electrosynthesis.
The balanced chemical equation for the conversion of potassium manganate(VII) to manganese(IV) dioxide is:
2KMnO4 → 2K+ + 2MnO2 + 3O2
From the equation, it can be seen that 2 moles of KMnO4 give 2 moles of MnO2. Therefore, 1 mole of KMnO4 will give 1 mole of MnO2, which is option A.
However, the formula used in this problem assumes that only one mole of KMnO4 is present, and therefore only 0.33 mole of MnO2 will be formed. This is because the reaction proceeds until all the KMnO4 is consumed, and the formula used calculates the amount of MnO2 formed when only 0.33 mole of KMnO4 is present.
Therefore, the correct answer to this problem is option D, 0.33 mole of MnO2.
In the electrosynthesis potassium manganate(VII) is converted to manga...
Manganese changes it's oxidation number
from +7 to +4. ∆=3e- gained
1F deposits or liberates 1 mole of e-
3F=from 1 mole of potassium (VII)manganate to form a mole of manganese (IV)oxide
1F of electrolysis=?Y
Y=1F×1 mole ÷3F
Y=0.33mole of manganese (IV)oxide