Question

question 15 (1 point)
how many water molecules must be added to the half-reaction to balance the mass of oxygen atoms?
cn⁻(aq) → cno⁻(s)
1 on the left side
1 on the right side
2 on the left side
2 on the right side
no water molecules are needed

Explanation:

Step1: Analyze oxygen atoms

On the left side (reactant side) of the half - reaction $\ce{CN^-(aq) -> CNO^-(s)}$, the reactant $\ce{CN^-}$ has 0 oxygen atoms. On the product side, $\ce{CNO^-}$ has 1 oxygen atom.

Step2: Determine water addition

To balance the oxygen atoms, we need to add a source of oxygen to the left - hand side. Water ($\ce{H_2O}$) contains one oxygen atom per molecule. So we add 1 water molecule to the left - hand side to provide the necessary oxygen atom. The half - reaction then becomes $\ce{CN^-(aq) + H_2O(l) -> CNO^-(s) + 2H^+(aq) + 2e^-}$ (we can also balance hydrogen and charge later, but for oxygen balance, 1 water on the left is needed).

Answer:

1 on the left side