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Electrochemistry: Question 2

Syllabus 6.1

Structured AS 8 marks

Sodium chlorate(V), NaClO3\text{NaClO}_3, is used industrially as a bleaching agent and, in dilute acidified solution, as a laboratory oxidising agent. When acidified sodium chlorate(V) solution is mixed with a solution of sodium sulfite, Na2SO3\text{Na}_2\text{SO}_3, the sulfite ions are oxidised to sulfate ions while the chlorate(V) ions are reduced to chloride ions.

(a) Determine the oxidation number of chlorine in ClO3−\text{ClO}_3^- and in Cl−\text{Cl}^-, and of sulfur in SO32−\text{SO}_3^{2-} and in SO42−\text{SO}_4^{2-}. [2]

(b) Using oxidation numbers, construct two separate balanced ionic half-equations, including H+(aq)\text{H}^+(aq) and H2O(l)\text{H}_2\text{O}(l) as appropriate: one for the reduction of ClO3−(aq)\text{ClO}_3^-(aq) to Cl−(aq)\text{Cl}^-(aq), and one for the oxidation of SO32−(aq)\text{SO}_3^{2-}(aq) to SO42−(aq)\text{SO}_4^{2-}(aq), both in acidic solution. [3]

(c) Combine your two half-equations from (b) to give the overall balanced ionic equation for the reaction, showing that both the atoms and the charge balance. [2]

(d) Identify, with a reason, the oxidising agent and the reducing agent in this reaction. [1]

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Worked solution

Part (a): Oxidation numbers

The rule for oxygen (in the absence of peroxides or fluorides) is oxidation number −2-2; the sum of oxidation numbers equals the overall ionic charge.

  • ClO3−\text{ClO}_3^-: three oxygens contribute 3(−2)=−63(-2) = -6; overall charge is −1-1, so x+(−6)=−1  ⟹  x=+5x + (-6) = -1 \implies x = +5.
  • Cl−\text{Cl}^-: a lone monatomic ion has an oxidation number equal to its own charge, so x=−1x = -1.
  • SO32−\text{SO}_3^{2-}: three oxygens contribute 3(−2)=−63(-2) = -6; overall charge is −2-2, so x+(−6)=−2  ⟹  x=+4x + (-6) = -2 \implies x = +4.
  • SO42−\text{SO}_4^{2-}: four oxygens contribute 4(−2)=−84(-2) = -8; overall charge is −2-2, so x+(−8)=−2  ⟹  x=+6x + (-8) = -2 \implies x = +6.

So chlorine falls from +5+5 to −1-1 (a change of 66), while sulfur rises from +4+4 to +6+6 (a change of 22).

Part (b): Constructing the two half-equations

Chlorine (reduction): the oxidation number falls by 66, so 66 electrons are gained per chlorine:

ClO3−(aq)+6e−→Cl−(aq)\text{ClO}_3^-(aq) + 6e^- \rightarrow \text{Cl}^-(aq)

Three oxygen atoms are lost on going from ClO3−\text{ClO}_3^- to Cl−\text{Cl}^-. In acidic solution these are removed as water, which requires H+(aq)\text{H}^+(aq) on the left:

ClO3−(aq)+6H+(aq)+6e−→Cl−(aq)+3H2O(l)\text{ClO}_3^-(aq) + 6\text{H}^+(aq) + 6e^- \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l)

Check: Atoms, Cl: 1=11 = 1 ✓. O: 3=33 = 3 (in 3H2O3\text{H}_2\text{O}) ✓. H: 6=66 = 6 (in 3H2O3\text{H}_2\text{O}) ✓. Charge, left: (−1)+6(+1)+6(−1)=−1(-1) + 6(+1) + 6(-1) = -1; right: (−1)+0=−1(-1) + 0 = -1. Equal ✓.

Sulfur (oxidation): the oxidation number rises by 22, so 22 electrons are lost per sulfur:

SO32−(aq)→SO42−(aq)+2e−\text{SO}_3^{2-}(aq) \rightarrow \text{SO}_4^{2-}(aq) + 2e^-

One extra oxygen atom is needed on going from SO32−\text{SO}_3^{2-} to SO42−\text{SO}_4^{2-}. In acidic solution this comes from water, releasing H+(aq)\text{H}^+(aq) on the right:

SO32−(aq)+H2O(l)→SO42−(aq)+2H+(aq)+2e−\text{SO}_3^{2-}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{SO}_4^{2-}(aq) + 2\text{H}^+(aq) + 2e^-

Check: Atoms, S: 1=11 = 1 ✓. O: 3+1=43 + 1 = 4 (in SO42−\text{SO}_4^{2-}) ✓. H: 2=22 = 2 (in 2H+2\text{H}^+) ✓. Charge, left: (−2)+0=−2(-2) + 0 = -2; right: (−2)+2(+1)+2(−1)=−2(-2) + 2(+1) + 2(-1) = -2. Equal ✓.

Part (c): Combining the half-equations

The chlorine half-equation transfers 66 electrons; the sulfur half-equation transfers only 22. To combine them, the electrons must match, so the sulfur half-equation is scaled by 33:

3SO32−(aq)+3H2O(l)→3SO42−(aq)+6H+(aq)+6e−3\text{SO}_3^{2-}(aq) + 3\text{H}_2\text{O}(l) \rightarrow 3\text{SO}_4^{2-}(aq) + 6\text{H}^+(aq) + 6e^-

Adding this to the chlorine half-equation:

ClO3−(aq)+6H+(aq)+6e−+3SO32−(aq)+3H2O(l)→Cl−(aq)+3H2O(l)+3SO42−(aq)+6H+(aq)+6e−\text{ClO}_3^-(aq) + 6\text{H}^+(aq) + 6e^- + 3\text{SO}_3^{2-}(aq) + 3\text{H}_2\text{O}(l) \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l) + 3\text{SO}_4^{2-}(aq) + 6\text{H}^+(aq) + 6e^-

The 6e−6e^- cancel, the 6H+(aq)6\text{H}^+(aq) cancel, and the 3H2O(l)3\text{H}_2\text{O}(l) cancel, leaving:

ClO3−(aq)+3SO32−(aq)→Cl−(aq)+3SO42−(aq)\text{ClO}_3^-(aq) + 3\text{SO}_3^{2-}(aq) \rightarrow \text{Cl}^-(aq) + 3\text{SO}_4^{2-}(aq)

Check (independent recomputation):

  • Atoms. Cl: 1=11 = 1 ✓. S: 3=33 = 3 ✓. O: left 3+3(3)=123 + 3(3) = 12; right 3(4)=123(4) = 12 ✓.
  • Charge, left: (−1)+3(−2)=−7(-1) + 3(-2) = -7; right: (−1)+3(−2)=−7(-1) + 3(-2) = -7. Equal ✓.

Both atoms and charge balance, confirming the overall ionic equation.

Part (d): Oxidising agent and reducing agent

ClO3−\text{ClO}_3^- is the oxidising agent: it is itself reduced (chlorine’s oxidation number falls from +5+5 to −1-1), and in doing so it removes electrons from, and so oxidises, the sulfite ions.

SO32−\text{SO}_3^{2-} is the reducing agent: it is itself oxidised (sulfur’s oxidation number rises from +4+4 to +6+6), and in doing so it supplies electrons to, and so reduces, the chlorate(V) ions.

Final answers

  • (a) Cl in ClO3−=+5\text{ClO}_3^- = +5; Cl in Cl−=−1\text{Cl}^- = -1; S in SO32−=+4\text{SO}_3^{2-} = +4; S in SO42−=+6\text{SO}_4^{2-} = +6.
  • (b) ClO3−(aq)+6H+(aq)+6e−→Cl−(aq)+3H2O(l)\text{ClO}_3^-(aq) + 6\text{H}^+(aq) + 6e^- \rightarrow \text{Cl}^-(aq) + 3\text{H}_2\text{O}(l); SO32−(aq)+H2O(l)→SO42−(aq)+2H+(aq)+2e−\text{SO}_3^{2-}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{SO}_4^{2-}(aq) + 2\text{H}^+(aq) + 2e^-
  • (c) ClO3−(aq)+3SO32−(aq)→Cl−(aq)+3SO42−(aq)\text{ClO}_3^-(aq) + 3\text{SO}_3^{2-}(aq) \rightarrow \text{Cl}^-(aq) + 3\text{SO}_4^{2-}(aq)
  • (d) Oxidising agent: ClO3−\text{ClO}_3^- (reduced). Reducing agent: SO32−\text{SO}_3^{2-} (oxidised).