Question

Write a balanced half-reaction for the reduction of aqueous arsenic acid H3AsO4 to gaseous arsine AsH3 in basic aqueous solution. Be sure to add physical state symbols where appropriate.

Answer 1

`H_3AsO_4(aq)+8H^+(aq)+8e^-=AsH_3(g)+4H_2O(l)`

Step-by-step explanation:

Answer 2

H₃AsO₄(aq) + 4 H₂O(l) + 8 e⁻ ⇒ AsH₃(g) + 8 OH⁻(aq)

Step-by-step explanation:

Let's consider the half-reaction for the reduction of aqueous arsenic acid to gaseous arsine in a basic aqueous solution.

H₃AsO₄(aq) ⇒ AsH₃(g)

We see that there is an excess of 4 oxygen atoms on the left side. So, we add 4 molecules of water to the left side and 8 hydroxyl ions to the right side.

H₃AsO₄(aq) + 4 H₂O(l) ⇒ AsH₃(g) + 8 OH⁻(aq)

We need to add 8 electrons to the left side to balance the reaction electrically.

H₃AsO₄(aq) + 4 H₂O(l) + 8 e⁻ ⇒ AsH₃(g) + 8 OH⁻(aq)