You perform an electrolysis of NaCl and collect one of the products in a test tube. You realize later you did not label the test tube. You dip a strip of damp pH paper into the test tube and get no reaction. You dip a dry strip of pH paper into the tube and there is no reaction. You hold a lit match near the mouth of the test tube and hear a loud 'pop'. Which product is in the test tube?
A. Chlorine gas
B. Sodium hydroxide (NaOH)
C. Hydrogen gas
D. Oxygen gas

Aqueous electrolysis of salt solutions finds the salt ions in competition with electrolysis of water at the respective electrodes. That is, water can undergo reduction at the cathode as can the cation of the salt; while at the anode water can undergo oxidation as can the anion of the salt. The reactions are as follows:

Cathode:

H₂O(l) + 2e⁻ ⇄ 2OH⁻(aq) + H₂(g)

Anode:

2H₂O(l) ⇄ O₂(g) + 4H⁺(aq) + 4e⁻

The distinction can be determined by comparing the 'Standard Reduction Potentials' of the half-reactions.

For NaCl(s) => Na⁺(aq) + Cl⁻(aq) :

At the Cathode:

Na⁺ + e⁻ => Na⁰(s); εo = -2.71v

2H₂O(l) + 2e⁻ ⇄ 2OH⁻(aq) + H₂(g); εo = -0.83v (Dominant Rxn at cathode – more positive value)

At the Anode:

Cl⁻(aq) => Cl₂(g) + 2e⁻; εo = 1.36v

2H₂O(l) => O₂(g) + 4H⁺ + 4e⁻; εo = 1.23v (Dominant Rxn at anode – more positive value)

Since the more dominant reaction at the cathode(reduction) produces hydroxide ions, the solution would become alkaline with Na⁺ ions already present.

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