Solve the redox equation in English

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To balance the redox equation $\text{Zn} + \text{BrO}_3^- \rightarrow \text{Zn}^{2+} + \text{Br}^-$ , we will use the half-reaction method in an acidic medium.

Separate the equation into half-reactions for oxidation and reduction.
Oxidation: $\text{Zn} \rightarrow \text{Zn}^{2+}$
Reduction: $\text{BrO}_3^- \rightarrow \text{Br}^-$
Balance atoms other than oxygen and hydrogen.
The zinc (Zn) and bromine (Br) atoms are already balanced in each half-reaction.
Balance oxygen (O) atoms by adding water molecules ( $\text{H}_2\text{O}$ ).
Oxidation: $\text{Zn} \rightarrow \text{Zn}^{2+}$ (no oxygen atoms)
Reduction: $\text{BrO}_3^- \rightarrow \text{Br}^- + 3\text{H}_2\text{O}$ (3 oxygen atoms on the left, so add 3 $\text{H}_2\text{O}$ on the right)
Balance hydrogen (H) atoms by adding hydrogen ions ( $\text{H}^+$ ).
Oxidation: $\text{Zn} \rightarrow \text{Zn}^{2+}$ (no hydrogen atoms)
Reduction: $6\text{H}^+ + \text{BrO}_3^- \rightarrow \text{Br}^- + 3\text{H}_2\text{O}$ (6 hydrogen atoms on the right in 3 $\text{H}_2\text{O}$ , so add 6 $\text{H}^+$ on the left)
Balance the charges by adding electrons ( $\text{e}^-$ ).
Oxidation:
Left side charge: 0
Right side charge: $+2$
To balance, add 2 electrons to the right:
$\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-$
Reduction:
Left side charge: $(+6) + (-1) = +5$
Right side charge: $(-1) + (0) = -1$
To balance, add 6 electrons to the left:
$6\text{e}^- + 6\text{H}^+ + \text{BrO}_3^- \rightarrow \text{Br}^- + 3\text{H}_2\text{O}$
Multiply the half-reactions to equalize the number of electrons.
The oxidation half-reaction has 2 electrons, and the reduction half-reaction has 6 electrons. Multiply the oxidation half-reaction by 3:
$3 \times (\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-) \Rightarrow 3\text{Zn} \rightarrow 3\text{Zn}^{2+} + 6\text{e}^-$
The reduction half-reaction remains unchanged:
$6\text{e}^- + 6\text{H}^+ + \text{BrO}_3^- \rightarrow \text{Br}^- + 3\text{H}_2\text{O}$
Add the half-reactions and cancel common terms.
$3\text{Zn} + (6\text{e}^- + 6\text{H}^+ + \text{BrO}_3^-) \rightarrow (3\text{Zn}^{2+} + 6\text{e}^-) + (\text{Br}^- + 3\text{H}_2\text{O})$
Cancel the 6 electrons from both sides to get the balanced equation:
$3\text{Zn} + 6\text{H}^+ + \text{BrO}_3^- \rightarrow 3\text{Zn}^{2+} + \text{Br}^- + 3\text{H}_2\text{O}$
Verify the balance of atoms and charges.
Atoms:
Zn: 3 on left, 3 on right (balanced)
Br: 1 on left, 1 on right (balanced)
O: 3 on left, 3 on right (balanced)
H: 6 on left, 6 on right (balanced)

Charges:
Left side: $0 + (+6) + (-1) = +5$
Right side: $3 \times (+2) + (-1) + 0 = +6 - 1 = +5$
The charges are balanced.

The balanced redox equation is:
$\boxed{3\text{Zn} + 6\text{H}^+ + \text{BrO}_3^- \rightarrow 3\text{Zn}^{2+} + \text{Br}^- + 3\text{H}_2\text{O}}$

Solve the redox equation in English

Step-by-step solution

Separate the equation into half-reactions for oxidation and reduction.

Balance atoms other than oxygen and hydrogen.

Balance oxygen (O) atoms by adding water molecules ( $\text{H}_2\text{O}$ ).

Balance hydrogen (H) atoms by adding hydrogen ions ( $\text{H}^+$ ).

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

Multiply the half-reactions to equalize the number of electrons.

Add the half-reactions and cancel common terms.

Verify the balance of atoms and charges.

Still stuck on this one?

More English Questions

Solve the redox equation in English

Step-by-step solution

Separate the equation into half-reactions for oxidation and reduction.

Balance atoms other than oxygen and hydrogen.

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

Balance hydrogen (H) atoms by adding hydrogen ions (H+\text{H}^+H+).

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

Multiply the half-reactions to equalize the number of electrons.

Add the half-reactions and cancel common terms.

Verify the balance of atoms and charges.

Still stuck on this one?

More English Questions

Balance oxygen (O) atoms by adding water molecules ( $\text{H}_2\text{O}$ ).

Balance hydrogen (H) atoms by adding hydrogen ions ( $\text{H}^+$ ).

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