Question 8 of 31intermediate🔧 ApplyShort Answer3 marks

Write the net ionic equation for the reaction of potassium dichromate(VI), K₂Cr₂O₇ with sodium sulphite, Na₂SO₃, in an acid solution to give chromium(III) ion and the sulphate ion.

Correct Answer

Cr2O72−(aq)+3SO32−(aq)+8H+(aq)→2Cr3+(aq)+3SO42−(aq)+4H2O(l)\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 3\text{SO}_3^{2-}(\text{aq}) + 8\text{H}^+ (\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq}) + 4\text{H}_2\text{O} (\text{l})

Exercise: Problem 7.8 | Q: Problem 7.8 | (Page Page 12)
For More Understanding

Explanation

To balance the redox reaction using the oxidation number method, we first write the skeletal ionic equation and assign oxidation numbers to identify the oxidant and reductant. The oxidation state of chromium decreases from +6 to +3, while the oxidation state of sulphur increases from +4 to +6. We equalize the increase and decrease in oxidation numbers, balance the charges by adding H+H^+ ions since it is in an acidic medium, and finally balance the hydrogen and oxygen atoms by adding water molecules.

Solution Steps

  1. Step 1: Write the skeletal ionic equation:

    Cr2O72−(aq)+SO32−(aq)→Cr3+(aq)+SO42−(aq)\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + \text{SO}_3^{2-}(\text{aq}) \rightarrow \text{Cr}^{3+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq})

  2. Step 2: Assign oxidation numbers and balance the change: The oxidation state of Cr changes from +6 to +3 (decrease of 3). The oxidation state of S changes from +4 to +6 (increase of 2). To make the increase and decrease equal, place 2 before Cr3+\text{Cr}^{3+} and 3 before SO32−\text{SO}_3^{2-} and SO42−\text{SO}_4^{2-}:

    Cr2O72−(aq)+3SO32−(aq)→2Cr3+(aq)+3SO42−(aq)\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 3\text{SO}_3^{2-}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq})

  3. Step 3: Balance the ionic charges: As the reaction occurs in an acidic medium, add 8H+8\text{H}^+ to the left side to equalize the charges:

    Cr2O72−(aq)+3SO32−(aq)+8H+→2Cr3+(aq)+3SO42−(aq)\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 3\text{SO}_3^{2-}(\text{aq}) + 8\text{H}^+ \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq})

  4. Step 4: Balance hydrogen and oxygen atoms: Add 4H2O4\text{H}_2\text{O} molecules to the right side to balance the hydrogen atoms:

    Cr2O72−(aq)+3SO32−(aq)+8H+(aq)→2Cr3+(aq)+3SO42−(aq)+4H2O(l)\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 3\text{SO}_3^{2-}(\text{aq}) + 8\text{H}^+ (\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{SO}_4^{2-}(\text{aq}) + 4\text{H}_2\text{O} (\text{l})
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