Balancing redox reactions: Examples

Now, let’s try balancing some sample redox reactions using the half equation method by applying the steps above in both an acidic, a basic and a neural dedium

Example in acidic medium

Balance the following redox reaction in acidic conditions.

\inline \fn_jvn \large {Cr_{2}O_{7}^{2-}}_{(aq)}+{HNO_{2}}_{(aq)}\rightleftharpoons {Cr^{3+}}_{(aq)}+{N0_{3}^{-}}_{(aq)}

Solution

Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.

$\inline \fn_jvn \large {Cr_{2}O_{7}^{2-}}_{(aq)}\rightarrow {Cr^{3+}}_{(aq)}$

$\inline \fn_jvn \large {HNO_{2}}_{(aq)}\rightarrow {NO_{3}^{-}}_{(aq)}$

Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:

\inline \fn_jvn \large {Cr_{2}O_{7}^{2-}}_{(aq)}\rightarrow {2Cr^{^{3+}}}_{(aq)}

$\inline \fn_jvn \large {HNO_{2}}_{(aq)}\rightarrow {NO_{3}^{-}}_{(aq)}$

Step 3: Add H₂O to balance oxygen. The chromium reaction needs to be balanced by adding 7 H₂O molecules. The other reaction also needs to be balanced by adding one water molecule. This yields:

$\inline \fn_jvn \large {Cr_{2}O_{7}^{2-}}_{(aq)}\rightarrow {2Cr^{3+}}_{(aq)}+{7H_{2}O}_{(l)}$

$\inline \fn_jvn \large {HNO_{2}}_{(aq)}+{H_{2}O}_{l}\rightarrow {NO_{3}^{-}}_{(aq)}$

{HNO}_{2}^{} (aq) + H_{2}^{} O (l) \to {NO}_{3}^{} - (aq)

Step 4: Balance hydrogen by adding protons (H⁺). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction.

\inline \fn_jvn \large {14H^{+}}_{(aq)}+{Cr_{2}O_{7}^{2-}}_{(aq)}\rightarrow {2Cr^{3+}}_{(aq)}+{7H_{2}O}_{(l)}

$\inline \fn_jvn \large {HNO_{2}}_{(aq)}+{H_{2}O}_{(l)}\rightarrow {3H^{+}}_{(aq)}+{NO_{3}^{-}}_{(aq)}$

Step 5: Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:

$\inline \fn_jvn \large 6e^{-}+{14H^{+}}_{(aq)}+{Cr_{2}O_{7}^{2-}}_{(aq)}\rightarrow {2Cr^{3+}}_{(aq)}+{7H_{2}O}_{_{(l)}}$

6 e^{-} + 14 H^{+} (aq) + {Cr}_{2}^{} O_{7}^{2 -} (aq) \to 2 {Cr}^{3 +} (aq) + 7 H_{2}^{} O (l)

\inline \fn_jvn \large {HNO_{2}}_{(aq)}+{H_{2}O}_{(l)}\rightarrow {3H^{+}}_{(aq)}+{NO_{3}^{-}}_{(aq)}+2e^{-}

Step 6: Scale the reactions so that the electrons are equal. The chromium reaction has 6e^– and the other reaction has 2e^–, so it should be multiplied by 3. This gives:

\inline \fn_jvn \large 3[{HNO_{2}}_{(aq)}+{H_{2}O}_{(l)}\rightarrow 3H^{+}_{(aq)}+{NO_{3}^{-}}_{(aq)}+2e^{-}]

$\inline \fn_jvn \large \Rightarrow {3HNO_{2}}_{(aq)}+{3H_{2}O}_{(l)}\rightarrow {9H^{+}}_{(aq)}+{3NO_{3}^{-}}_{(aq)}+6e^{-}$

$\inline \fn_jvn \large 6e^{-}+{14H^{+}}_{(aq)}+{Cr_{2}O_{7}^{2-}}_{(aq)}\rightarrow {2Cr^{3+}}_{(aq)}+{7H_{2}O}_{(l)}$

Step 7: Add the reactions and cancel out common terms.

$\inline \fn_jvn \large {3HNO_2}_{(aq)} + {5H^+}_{(aq)} + {Cr_2O_7^{2-}}_{(aq)} \rightleftharpoons {3NO_3^-}_{(aq)} + {2Cr^{3+}}_{(aq)} + {4H_2O}_{(l)}$