Write a balanced half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution. Be sure to add physical state symbols where appropriate.

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)

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.