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