Answer:
2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)
Explanation:
Let's consider the unbalanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution.
H₂O(l) ⇒ H₂(g)
First, we will perform the mass balance. We will balance oxygen atoms by multiplying H₂O by 2 and adding 2 OH⁻ to the right side.
2 H₂O(l) ⇒ H₂(g) + 2 OH⁻(aq)
Then, we perform the charge balance by adding 2 electrons to the left side.
2 H₂O(l) + 2 e⁻ ⇒ H₂(g) + 2 OH⁻(aq)
