Question 21.9: Predicting the Electrolysis Products of a Molten Salt Mixtur......
Predicting the Electrolysis Products of a Molten Salt Mixture
Problem A chemical engineer melts a naturally occurring mixture of NaBr and MgCl_2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced half-reactions and the overall cell reaction.
Plan We have to determine which metal and nonmetal will form more easily at each electrode. We list the ions as oxidizing or reducing agents and then use periodic trends to determine which metal ion is more easily reduced and which nonmetal ion more easily oxidized.
Solution Listing the ions as oxidizing or reducing agents:
The possible oxidizing agents are Na^+ and Mg^{2+}; these cations may undergo
reduction.
The possible reducing agents are Br^− and Cl^−; these anions may undergo oxidation.
Determining the cathode product (more easily reduced cation): Mg is to the right of Na in Period 3. IE increases from left to right, so it is harder to remove e^− from Mg. Thus, Mg^{2+} has a greater attraction for e^− and is more easily reduced (stronger oxidizing agent). Mg^{2+} will be reduced preferentially at the cathode:
Mg^{2+}(l) + 2e^− ⟶ Mg(l) \text{ [cathode; reduction]}Determining the anode product (more easily oxidized anion): Br is below Cl in Group 7A(17). EN decreases down the group, so Br accepts e^− less readily. Thus, Br^− loses its e^− more easily and is more easily oxidized (stronger reducing agent). Br^− will be oxidized preferentially at the anode:
2Br^−(l) ⟶ Br_2(g) + 2e^− \text{ [anode; oxidation]}Writing the overall cell reaction:
Mg^{2+}(l) + 2Br^−(l) ⟶ Mg(l) + Br_2(g) \text{ [overall]}Comment The cell temperature must be high enough to keep the salt mixture molten. In this case, the temperature is greater than the melting point of Mg, so it appears as a liquid in the equation, and greater than the boiling point of Br_2, so it appears as a gas.