Question

Write balanced net ionic equations for the following reactions in basic medium. (a) \(\mathrm{Ca}(s)+\mathrm{VO}_{4}^{3-}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{V}^{2+}(a q)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{BiO}_{3}^{-}(a q) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{Bi}^{3+}(a q)\) (c) \(\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{O}_{2}(g)+\mathrm{Pb}^{2+}\) (b) \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)+\mathrm{I}_{3}^{-}(a q)\)

Short answer

Question: Write balanced net ionic equations for the following reactions that occur under basic conditions: (a) Ca(s) + VO4^3-(aq) → Ca^2+(aq) + V^2+(aq) (b) C2H4(g) + BiO3^-(aq) → CO2(g) + Bi^3+(aq) (c) PbO2(s) + H2O → O2(g) + Pb^2+(aq) (d) IO3^-(aq) + Cl^-(aq) → I3^-(aq) + Cl2(g) Answer: (a) Ca(s) + VO4^3-(aq) + 4OH^-(aq) → Ca^2+(aq) + V^2+(aq) + 4OH^-(aq) (b) 3C2H4(g) + 8BiO3^-(aq) + 24OH^-(aq) → 6CO2(g) + 8Bi^3+(aq) + 24OH^-(aq) (c) PbO2(s) + 2H2O + 4OH^-(aq) → O2(g) + Pb^2+(aq) + 4OH^-(aq) (d) IO3^-(aq) + 6Cl^-(aq) + 6OH^-(aq) → I3^-(aq) + 3Cl2(g) + 3H2O

Step-by-step solution

Identify the Redox Half-Reactions

In the reaction, Ca(s) undergoes oxidation, so the half-reaction for this is: $$\mathrm{Ca}(s) \longrightarrow \mathrm{Ca}^{2+}(a q) + 2e^-$$ Also, VO4^3-(aq) undergoes reduction, so the half-reaction for this is: $$\mathrm{VO}_{4}^{3-}(a q) + 2e^- \longrightarrow \mathrm{V}^{2+}(a q)+ 4\mathrm{OH}^{-}(aq)$$ We added 4OHˉ on the right side to balance the oxygens (basic medium).

Multiply the Half-Reactions and Add Them Up

There's no need to multiply the half-reactions since they both involve two electrons. Now, add them up: $$\mathrm{Ca}(s)+\mathrm{VO}_{4}^{3-}(a q) + 4\mathrm{OH}^{-}(aq) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{V}^{2+}(a q) + 4\mathrm{OH}^{-}(aq)$$ (b)

Identify the Redox Half-Reactions

For this reaction, the oxidation half-reaction is (given that one C undergoes oxidation to CO2): $$\mathrm{C}_{2} \mathrm{H}_{4}(g) \longrightarrow 2\mathrm{CO}_{2}(g) + 8e^-$$ And the reduction half-reaction is: $$\mathrm{BiO}_{3}^{-}(a q) + 3e^- \longrightarrow \mathrm{Bi}^{3+}(a q) + 3\mathrm{OH}^{-}(a q)$$

Multiply the Half-Reactions and Add Them Up

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 8 and then add them up: $$3\mathrm{C}_{2} \mathrm{H}_{4}(g) + 8\mathrm{BiO}_{3}^{-}(a q) + 24\mathrm{OH}^{-}(a q) \longrightarrow 6\mathrm{CO}_{2}(g) + 8\mathrm{Bi}^{3+}(a q) + 24\mathrm{OH}^{-}(a q)$$ (c)

Identify the Redox Half-Reactions

For this reaction, the half-reaction for the reduction of water is: $$\mathrm{H}_{2} \mathrm{O} + 2e^- \longrightarrow 2\mathrm{OH}^{-}(a q)$$ And the oxidation half-reaction is: $$\mathrm{PbO}_{2}(s) \rightarrow \mathrm{O}_{2}(g)+\mathrm{Pb}^{2+} + 4e^-$$

Multiply the Half-Reactions and Add Them Up

Multiply the reduction half-reaction by 2 and leave the oxidation half-reaction as-is. Then, add up the two half-reactions: $$\mathrm{PbO}_{2}(s)+2\mathrm{H}_{2} \mathrm{O} + 4\mathrm{OH}^{-}(a q)\longrightarrow \mathrm{O}_{2}(g)+\mathrm{Pb}^{2+} + 4\mathrm{OH}^{-}(a q)$$ (d)

Identify the Redox Half-Reactions

For this reaction, the reduction half-reaction is: $$\mathrm{IO}_{3}^{-}(a q) + 6e^- + 6\mathrm{OH}^{-}(a q)\longrightarrow I_3^{-}(a q) + 3\mathrm{H}_2\mathrm{O}$$ And the oxidation half-reaction is: $$2\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g) + 2e^-$$

Multiply the Half-Reactions and Add Them Up

Multiply the reduction half-reaction by 1 and the oxidation half-reaction by 3, and add them up: $$\mathrm{IO}_{3}^{-}(a q) + 6\mathrm{Cl}^{-}(a q) + 6\mathrm{OH}^{-}(a q) \longrightarrow I_3^{-}(a q) + 3\mathrm{Cl}_{2}(g) + 3\mathrm{H}_2\mathrm{O}$$

Key concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are processes where electrons are transferred between two substances. In these reactions, one species is oxidized, meaning it loses electrons, and another is reduced, meaning it gains electrons. Understanding the concept of oxidation states is key to identifying and balancing redox reactions. An increase in oxidation state indicates oxidation, while a decrease signifies reduction.

For example, in the reaction between calcium (Ca) and vanadate (VO₄^3-), Ca goes from an oxidation state of 0 to +2, losing two electrons (oxidized), and VO₄^3- gains electrons to become V²⁺ (reduced). Recognizing these changes is the first step in writing net ionic equations for such reactions.

Chemical Reaction Balancing

Balancing chemical reactions is essential to convey the conservation of mass and charge. In a balanced equation, the number of atoms for each element and the overall charge must be the same on both the reactants and products side. This can be particularly challenging in complex redox reactions, especially in a basic medium where hydroxide ions (OH^-) complicate the process.

To accurately balance these reactions, it is important to adjust the coefficients of the reactants and products. In the given examples, different strategies are applied, like adding water molecules to balance oxygen atoms or hydroxide ions to balance charge in a basic solution. Always ensure that the total charge and the total number of atoms for each element are balanced before declaring the equation complete.

Half-Reaction Method

The half-reaction method is a systematic approach to balancing redox reactions, particularly useful in both acidic and basic mediums. This method involves separating the overall chemical reaction into two half-reactions: one for oxidation and one for reduction. Each half-reaction is balanced for mass and charge independently, often by adding electrons, water, and hydroxide ions as necessary.

Once the half-reactions are balanced, they are then recombined by multiplying them by appropriate factors so that the number of electrons lost in the oxidation reaction equals the number of electrons gained in the reduction reaction. The final step is to combine these half-reactions, canceling out electrons and any other species that appear on both sides of the equation to present a balanced net ionic equation. This method ensures a thorough and methodical approach to balancing, particularly in complex scenarios.