Question

balance the following equations by the half-reaction method.
a. Fe + HCl = HFeCl4 + H2

Answer 1

Answer:

2Fe + 8HCl ⟶ 2HFeCl₄ + 3H₂  

Explanation:

1. Write the skeleton equation

Fe + HCl ⟶ HFeCl₄ + H₂

2. Separate it into two half-reactions.

 Fe ⟶ HFeCl₄

HCl ⟶ H₂

3. Balance all atoms other than H and O

Balance Cl by adding Cl⁻.

Fe + 4Cl⁻ ⟶ HFeCl₄

        HCl ⟶ H₂ + Cl⁻

4. Balance H by adding H⁺ ions to the deficient side.

H⁺ + Fe + 4Cl⁻ ⟶ HFeCl₄

        HCl + H⁺ ⟶ H₂ + Cl⁻

5. Balance charge by adding electrons to the deficient side.

H⁺ + Fe + 4Cl⁻ ⟶ HFeCl₄ + 3e⁻

HCl + H⁺ + 2e⁻ ⟶ H₂ + Cl⁻

6. Multiply each half-reaction by a number to equalize the electrons transferred.

2 × [H⁺ + Fe + 4Cl⁻ ⟶ HFeCl₄ + 3e⁻]

3 × [HCl + H⁺ + 2e⁻ ⟶ H₂ + Cl⁻]

7. Add the two half-reactions.

                     2H⁺ + 2Fe + 8Cl⁻ ⟶ 2HFeCl₄ + 6e⁻

                     3HCl + 3H⁺ + 6e⁻ ⟶ 3H₂ + 3Cl⁻                            

2Fe + 3HCl + 5H⁺ + 8Cl⁻ + 6e⁻ ⟶ 2HFeCl₄ + 3H₂ + 3Cl⁻ + 6e⁻

8. Cancel species that occur on each side of the equation

2Fe + 3HCl + 5H⁺ + 8Cl⁻ + 6e⁻ ⟶ 2HFeCl₄ + 3H₂ + 3Cl⁻ + 6e⁻

becomes

2Fe + 3HCl + 5H⁺ + 5Cl⁻ ⟶ 2HFeCl₄ + 3H₂

9. Combine the H⁺ and Cl⁻ ions

2Fe + 3HCl + 5HCl ⟶ 2HFeCl₄ + 3H₂

becomes

2Fe + 8HCl ⟶ 2HFeCl₄ + 3H₂

10. Check that all atoms are balanced.

$\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{Fe} & 1 & 1\\\text{H} & 8 & 8\\\text{Cl} & 8 & 8\\\end{array}$

Everything checks. The balanced equation is

2Fe + 8HCl ⟶ 2HFeCl₄ + 3H₂

Answer 2

The balanced redox equation is as a follows:

• 2Fe + 8HCl ⟶ 2HFeCl₄ + 3H₂

What is the half-reaction method of balancing redox equations?

The half-reaction method of balancing redox equations involves splitting the redox equation into oxidation and reduction half-equations and then balancing atoms and charges of both half equations.

The equation of the redox reaction is given below:

• Fe + HCl ⟶ HFeCl₄ + H₂

The half-reactions are as follows:

• Oxidation- Fe ⟶ HFeCl₄
• Reduction- HCl ⟶ H₂

Balancing the atoms of the half-reaction:

Fe + 4Cl⁻ + H⁺ + ⟶ HFeCl₄

HCl + H⁺ ⟶ H₂ + Cl⁻

Balancing the charge by adding electrons to the deficient side.

H⁺ + Fe + 4Cl⁻ ⟶ HFeCl₄ + 3e⁻

HCl + H⁺ + 2e⁻ ⟶ H₂ + Cl⁻

Multiply oxidation half-reaction by 2 and reduction half-reaction by 3 to balance the electrons transferred.

2 H⁺ + 2 Fe + 8 Cl⁻ ⟶ 2 HFeCl₄ + 6 e⁻

3 HCl + 3 H⁺ + 6e⁻ ⟶ 3 H₂ + 3Cl⁻

Combine the two half-reactions and write the net ionic equation by cancelling electrons and spectator ions.

2Fe + 8HCl ⟶ 2HFeCl₄ + 3H₂

Therefore, the balanced equation of the reaction is; 2Fe + 8HCl ⟶ 2HFeCl₄ + 3H₂

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