213k views
2 votes
G. Amount of charge required to reduce

Imole of NO3-to N2O is
a) 3F
c) 4F
b) 8F
d) 9F

2 Answers

2 votes
the answer would be c 4f
User Eugstman
by
4.7k points
1 vote

Answer:


\boxed{\text{c) 4 F}}

Step-by-step explanation:

1. Write the skeleton equation for the half-reaction

NO₃⁻ ⟶ N₂O

2. Balance all atoms other than H and O

2NO₃⁻ ⟶ N₂O

3. Balance O by adding H₂O molecules to the deficient side.

2NO₃⁻ ⟶ N₂O + 5H₂O

4. Balance H by adding H⁺ ions to the deficient side.

2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O

5. Balance charge by adding electrons to the deficient side.

2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O

The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F


\text{The amount of charge required to reduce 1 mol of NO$_(3)^(-)$ is \boxed{\textbf{4 F}}}

User Taras Vovkovych
by
5.3k points