You run 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 it turn dark blue. you hold a lit match near the mouth of the test tube and get no reaction. which product is in the test tube? a. sodium hydroxide (naoh) b. chlorine gas c. solid nacl d. hydrogen gas

Answer = 'A' => Sodium Hydroxide (NaOH)

Explanation:

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 competing reactions are as follows:

Cathode:

Cation + electrons => Reduced form of metal in Basic Standard State

H₂O(l) + 2e⁻ ⇄ 2OH⁻(aq) + H₂(g)

Anode:  

Anion ⇄ Basic Standard State element + electrons

2H₂O(l) ⇄ O₂(g) + 4H⁺(aq) + 4e⁻

The distinction can be determined by comparing the 'Standard Reduction Potentials' of the half-reactions. At the cathode (site of reduction), the more positive electrode potential will dominate while at the anode (site of oxidation) the more negative electrode potential will dominate.  

For NaCl(s) => Na⁺(aq) + Cl⁻(aq) finds the ions in competition with the oxidation and reduction reactions of water.  

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 produces hydroxide ions, the solution would become alkaline with Na⁺ ions already present. NaOH will effect the change in color of red litmus to blue. The gases produced in the half-reactions would have evaporated and not be present for flame test.