Question

Balance the following half-equations. Balance (a) and (b) in acidic medium, \((\mathrm{c})\) and \((\mathrm{d})\) in basic medium. (a) \(\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow \mathrm{CO}_{2}(g)\) (b) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NH}_{4}^{+}(a q)\) (c) \(\mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{Fe}(s)\) (d) \(\mathrm{V}^{2+}(a q) \longrightarrow \mathrm{VO}_{3}^{-}(a q)\)

Short answer

Question: Provide the balanced half-equation for the following reactions: (a) CH3OH(aq) ⟶ CO2(g) (Acidic medium) (b) NO3^-(aq) ⟶ NH4^+(aq) (Acidic medium) (c) Fe^3+(aq) ⟶ Fe(s) (Basic medium) (d) V^2+(aq) ⟶ VO3^-(aq) (Basic medium) Answer: (a) CH3OH(aq) + 6H+(aq) + 6e- ⟶ CO2(g) + 6H2O(l) (b) NO3^-(aq) + 10H+(aq) + 8e- ⟶ NH4^+(aq) + 3H2O(l) (c) Fe^3+(aq) + 3e- ⟶ Fe(s) (d) 2V^2+(aq) + 10H2O(l) ⟶ V2O6^2-(aq) + 20OH^-(aq) + 6e-

Step-by-step solution

Balancing Elements other than Hydrogen and Oxygen

For all half-equations, first, we balance the elements other than Hydrogen and Oxygen on both sides. For equation (a), C and O are balanced. For equation (b), N is balanced. For equation (c), Fe is balanced. For equation (d), V is balanced. Now we can move on to balancing Hydrogen and Oxygen.

Balancing Oxygen with Water Molecules

We will balance the O atoms by adding H2O molecules to the side that needs more O atoms. For (a): \(\mathrm{CH}_3\mathrm{OH}(aq) \longrightarrow \mathrm{CO}_2(g) + \mathrm{H}_2\mathrm{O}(l)\) For (b): \(\mathrm{NO}_3^-(aq)\) ⟶ \(\mathrm{NH}_4^+(aq) + 3\mathrm{H}_2\mathrm{O}(l)\) For (c) and (d), O atoms are already balanced.

Balancing Hydrogen with H+ ions (For acidic medium)

For equations (a) and (b), we will balance the H atoms by adding H+ ions. For (a): \(\mathrm{CH}_3\mathrm{OH}(aq) + 6\mathrm{H}^+(aq) \longrightarrow \mathrm{CO}_2(g) + \mathrm{H}_2\mathrm{O}(l)\) For (b): \(\mathrm{NO}_3^-(aq) + 10\mathrm{H}^+(aq) \longrightarrow \mathrm{NH}_4^+(aq) + 3\mathrm{H}_2\mathrm{O}(l)\)

Balancing Hydrogen with H2O and OH- ions (For basic medium)

For equations (c) and (d), we will balance the H atoms by adding H2O to the side needing H atoms and then adding the same amount of OH- ions to the other side. For (c): \(\mathrm{Fe}^{3+}(aq) + 3\mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)\) (There are no H atoms in this equation; therefore, the equation is already balanced.) For (d): \(\mathrm{V}^{2+}(aq) + 5\mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{VO}_3^-(aq) + 10\mathrm{OH}^-(aq)\)

Balancing the Charge with Electrons

We will balance the charge on both sides of the equation by adding electrons (e-). For (a): \(\mathrm{CH}_3\mathrm{OH}(aq) + 6\mathrm{H}^+(aq) + 6\mathrm{e}^{-} \longrightarrow \mathrm{CO}_2(g) + 6\mathrm{H}_2\mathrm{O}(l)\) For (b): \(\mathrm{NO}_3^-(aq) + 10\mathrm{H}^+(aq) + 8\mathrm{e}^{-} \longrightarrow \mathrm{NH}_4^+(aq) + 3\mathrm{H}_2\mathrm{O}(l)\) For (c): Equation (c) is already balanced in terms of charge, and there is no need for adding electrons. For (d): 2\(\mathrm{V}^{2+}(aq) + 10\mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{V}_2\mathrm{O}_6^{2-}(aq) + 20\mathrm{OH}^-(aq) + 6\mathrm{e}^{-}\) So, the balanced half-equations are: (a) \(\mathrm{CH}_3\mathrm{OH}(aq) + 6\mathrm{H}^+(aq) + 6\mathrm{e}^{-} \longrightarrow \mathrm{CO}_2(g) + 6\mathrm{H}_2\mathrm{O}(l)\) (b) \(\mathrm{NO}_3^-(aq) + 10\mathrm{H}^+(aq) + 8\mathrm{e}^{-} \longrightarrow \mathrm{NH}_4^+(aq) + 3\mathrm{H}_2\mathrm{O}(l)\) (c) \(\mathrm{Fe}^{3+}(aq) + 3\mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)\) (d) 2\(\mathrm{V}^{2+}(aq) + 10\mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{V}_2\mathrm{O}_6^{2-}(aq) + 20\mathrm{OH}^-(aq) + 6\mathrm{e}^{-}\)