Balance the redox reaction by Half reaction method.
${F e}^{2 +} (a q) + {C r}_{2} O_{7}^{2 -} (a q) (acid medium) - ------- \to {F e}^{3 +} (a q) + {C r}^{3 +} (a q)$

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Solution

Step 1: Write the unbalanced ionic equation,
$F e^{2 +} (a q) + C r_{2} O_{7}^{2 -} (a q) ⟶ F e^{3 +} (a q) + C r^{3 +} (a q)$
Step 2: Write separate half-reactions for the oxidation and the reduction processes. Determine the oxidation numbers first, if necessary,
$F e^{+ 2} \to F e^{+ 3}$
$C r^{+ 6} \to C r^{+ 3}$
step 3: Balance the atoms in the half-reactions other than hydrogen and oxygen. In the oxidation half-reaction above, the iron atoms are already balanced. The reduction half-reaction needs to be balanced with the chromium atoms,
Step 4: Balance oxygen atoms by adding water molecules to the appropriate side of the equation. For the reduction half-reaction above, seven $H_{2} O$ molecules will be added to the product side,
$C r_{2} O_{7}^{2 -} (a q) \to 2 C r^{3 +} (a q) + 7 H_{2} O (l)$
Now the hydrogen atoms need to be balanced. In an acidic medium, add hydrogen ions to balance. In this example, fourteen $H^{+}$ ions will be added to the reactant side.
Step 5: Balance the charges by adding the electrons,
$6 F e^{2 +} (a q) + C r_{2} O_{7}^{2 -} (a q) ⟶ 6 F e^{3 +} (a q) + 2 C r^{3 +} (a q) + 7 H_{2} O$ .

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