Question

sodium hydroxide can be produced commercially from the electrolysis of sodium chloride solutions, as indicated by the following balanced equation.
2 nacl(aq) + 2 h₂o(l) → 2 naoh(aq) + cl₂(g) + h₂(g)
δg°_rxn = 400 kj/mol
20 mark for review
which of the following is the best evidence that the reaction is an oxidation-reduction reaction?
a heat is consumed during the reaction.
b the oxidation numbers of both na and o change as the reaction proceeds.
c the oxidation number of each cl atom changes from -1 to 0, and some h atoms change from +1 to 0.
d each cl atom gains an electron, and each h₂o molecule loses a proton.

Explanation:

Step1: Analyze oxidation - reduction reactions

Oxidation - reduction reactions involve changes in oxidation numbers. In the reaction \(2NaCl(aq)+2H_{2}O(l)\to2NaOH(aq)+Cl_{2}(g)+H_{2}(g)\), for chlorine:
In \(NaCl\), the oxidation number of \(Cl\) is \(- 1\). In \(Cl_{2}\), the oxidation number of \(Cl\) is \(0\). For hydrogen:
In \(H_{2}O\), the oxidation number of \(H\) is \(+1\). In \(H_{2}\), the oxidation number of \(H\) is \(0\).

Step2: Check each option

• Option A: Heat consumption does not indicate an oxidation - reduction reaction. Many non - redox reactions also consume heat.
• Option B: The oxidation number of \(Na\) is \(+1\) in both \(NaCl\) and \(NaOH\), and the oxidation number of \(O\) is \(-2\) in both \(H_{2}O\) and \(NaOH\). So, this option is incorrect.
• Option C: The oxidation number of \(Cl\) changes from \(-1\) (in \(NaCl\)) to \(0\) (in \(Cl_{2}\)), and the oxidation number of \(H\) changes from \(+1\) (in \(H_{2}O\)) to \(0\) (in \(H_{2}\)). This shows a transfer of electrons (oxidation - reduction).
• Option D: In \(H_{2}O\), \(H\) is \(+1\) and \(O\) is \(-2\). There is no loss of a proton from \(H_{2}O\) in the context of redox (proton loss is more related to acid - base in some cases, not redox here).

Answer:

C. The oxidation number of each \(Cl\) atom changes from \(-1\) to \(0\), and some \(H\) atoms change from \(+1\) to \(0\)