Answer:
Reduction: 2 H⁺(aq) + H₂O₂(aq) + 2 e⁻ ⇒ 2 H₂O(l)
Oxidation: H₂O₂(aq) ⇒ O₂(g) + 2 H⁺(aq) + 2 e⁻
Explanation:
In H₂O₂, hydrogen has the oxidation number +1 and oxygen the oxidation number -1.
In the reduction half-reaction (H₂O₂ is the oxidizing agent), H₂O₂ forms H₂O. The oxidation number of oxygen decreases from -1 to -2.
2 H⁺(aq) + H₂O₂(aq) + 2 e⁻ ⇒ 2 H₂O(l)
In the oxidation half-reduction (H₂O₂ is the reducing agent), H₂O₂ forms O₂. The oxidation number of oxygen increases from -1 to 0.
H₂O₂(aq) ⇒ O₂(g) + 2 H⁺(aq) + 2 e⁻
The answer is that all of them are correct.
Given:
rate = k [A]2
concentration is
0.10 moles/liter
rate is 2.7 × 10-5 M*s-1
Required:
Value of k
Solution:
rate = k [A]2
2.7 × 10-5 M*s-1
= k (0.10 moles/liter)^2
k = 2.7 x 10^-3
liter per mole per second
Ca + Cl2 = CaCl₂
A synthesis <span>reaction.</span>