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Sagot :
Solution:
Step 1: Separate the equation into two half-reactions.
Oxidation: [tex]\text{Cl}^{-} → \text{Cl}_{2}[/tex]
Reduction: [tex]\text{Cr}_{2}\text{O}_{7}^{2-} → \text{Cr}^{3+}[/tex]
Step 2: Balance each half-reaction for number and type of atoms and charges. For reactions in an acidic medium, add [tex]H_{2}O[/tex] to balance the O atoms and add [tex]H^{+}[/tex] to balance the H atoms.
For oxidation half-reaction:
[tex]\text{Cl}^{-} → \text{Cl}_{2}[/tex]
[tex]2\text{Cl}^{-} → \text{Cl}_{2}[/tex]
[tex]2\text{Cl}^{-} → \text{Cl}_{2} + 2e^{-}[/tex]
For reduction half-reaction:
[tex]\text{Cr}_{2}\text{O}_{7}^{2-} → \text{Cr}^{3+}[/tex]
[tex]\text{Cr}_{2}\text{O}_{7}^{2-} → 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}[/tex]
[tex]14\text{H}^{+} + \text{Cr}_{2}\text{O}_{7}^{2-} → 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}[/tex]
[tex]14\text{H}^{+} + \text{Cr}_{2}\text{O}_{7}^{2-} + 6e^{-} → 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}[/tex]
Step 3: Multiply both half-reactions by a certain number to equalize the number of electrons.
For oxidation half-reaction:
[tex]2\text{Cl}^{-} → \text{Cl}_{2} + 2e^{-}[/tex]
[tex](2\text{Cl}^{-} → \text{Cl}_{2} + 2e^{-}) × 3[/tex]
[tex]6\text{Cl}^{-} → 3\text{Cl}_{2} + 6e^{-}[/tex]
For reduction half-reaction:
[tex]14\text{H}^{+} + \text{Cr}_{2}\text{O}_{7}^{2-} + 6e^{-} → 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}[/tex]
[tex](14\text{H}^{+} + \text{Cr}_{2}\text{O}_{7}^{2-} + 6e^{-} → 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}) × 1[/tex]
[tex]14\text{H}^{+} + \text{Cr}_{2}\text{O}_{7}^{2-} + 6e^{-} → 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}[/tex]
Step 4: Add the two half-reactions.
[tex]6\text{Cl}^{-} → 3\text{Cl}_{2} + 6e^{-}[/tex]
[tex]\underline{14\text{H}^{+} + \text{Cr}_{2}\text{O}_{7}^{2-} + 6e^{-} → 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O}}[/tex]
[tex]\boxed{14\text{H}^{+} + \text{Cr}_{2}\text{O}_{7}^{2-} + 6\text{Cl}^{-} → 2\text{Cr}^{3+} + 3\text{Cl}_{2} + 7\text{H}_{2}\text{O}}[/tex]
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