The unbalanced reaction is:

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Here's how to balance the redox reaction in acidic medium:

The unbalanced reaction is:

\text{Fe} + \text{ClO}_3^- \rightarrow \text{Fe}^{3+} + \text{Cl}^-

Separate the reaction into half-reactions.
Oxidation half-reaction:
$\text{Fe} \rightarrow \text{Fe}^{3+}$
Reduction half-reaction:
$\text{ClO}_3^- \rightarrow \text{Cl}^-$
Balance atoms other than oxygen and hydrogen.
Both Fe and Cl atoms are already balanced in their respective half-reactions.
Balance oxygen atoms by adding $\text{H}_2\text{O}$ .
Oxidation:
$\text{Fe} \rightarrow \text{Fe}^{3+}$
Reduction:
$\text{ClO}_3^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}$
Balance hydrogen atoms by adding $\text{H}^+$ (since it's an acidic medium).
Oxidation:
$\text{Fe} \rightarrow \text{Fe}^{3+}$
Reduction:
$6\text{H}^+ + \text{ClO}_3^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}$
Balance the charge by adding electrons ( $\text{e}^-$ ).
Oxidation:
$\text{Fe} \rightarrow \text{Fe}^{3+} + 3\text{e}^-$
Reduction:
$6\text{H}^+ + \text{ClO}_3^- + 6\text{e}^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}$
Make the number of electrons equal in both half-reactions.
Multiply the oxidation half-reaction by 2 to get 6 electrons:
$2(\text{Fe} \rightarrow \text{Fe}^{3+} + 3\text{e}^-)$ $2\text{Fe} \rightarrow 2\text{Fe}^{3+} + 6\text{e}^-$
The reduction half-reaction remains:
$6\text{H}^+ + \text{ClO}_3^- + 6\text{e}^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}$
Add the two balanced half-reactions and cancel out common species (electrons).
$2\text{Fe} + 6\text{H}^+ + \text{ClO}_3^- + 6\text{e}^- \rightarrow 2\text{Fe}^{3+} + 6\text{e}^- + \text{Cl}^- + 3\text{H}_2\text{O}$
Write the final balanced equation.
$2\text{Fe} + \text{ClO}_3^- + 6\text{H}^+ \rightarrow 2\text{Fe}^{3+} + \text{Cl}^- + 3\text{H}_2\text{O}$
The balanced equation is:
$\boxed{2\text{Fe} + \text{ClO}_3^- + 6\text{H}^+ \rightarrow 2\text{Fe}^{3+} + \text{Cl}^- + 3\text{H}_2\text{O}}$

The unbalanced reaction is:

Step-by-step solution

Separate the reaction into half-reactions.

Balance atoms other than oxygen and hydrogen.

Balance oxygen atoms by adding $\text{H}_2\text{O}$ .

Balance hydrogen atoms by adding $\text{H}^+$ (since it's an acidic medium).

Balance the charge by adding electrons ( $\text{e}^-$ ).

Make the number of electrons equal in both half-reactions.

Add the two balanced half-reactions and cancel out common species (electrons).

Write the final balanced equation.

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More Chemistry Questions

The unbalanced reaction is:

Step-by-step solution

Separate the reaction into half-reactions.

Balance atoms other than oxygen and hydrogen.

Balance oxygen atoms by adding H2O\text{H}_2\text{O}H2​O.

Balance hydrogen atoms by adding H+\text{H}^+H+ (since it's an acidic medium).

Balance the charge by adding electrons (e−\text{e}^-e−).

Make the number of electrons equal in both half-reactions.

Add the two balanced half-reactions and cancel out common species (electrons).

Write the final balanced equation.

Still stuck on this one?

More Chemistry Questions

Balance oxygen atoms by adding $\text{H}_2\text{O}$ .

Balance hydrogen atoms by adding $\text{H}^+$ (since it's an acidic medium).

Balance the charge by adding electrons ( $\text{e}^-$ ).