Question

Current is applied to a molten mixture of AgF , FeCl2 , and AlBr3 . What is produced at each electrode? STRATEGY Rank the cations by reduction potential. The ion with the highest reduction potential is favored for reduction. Rank anions by oxidation potential. The ion with the highest oxidation potential is favored for oxidation. Determine the products of the favored half‑reactions. Step 1: The cations Ag+ , Fe2+ , and Al3+ will all compete for reduction at the cathode.

Answer 1

Answer:

Cathode: Ag

Anode: Br₂

Explanation:

In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:

Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V

Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V

Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V

As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.

For the anode an oxidation must occurs, so the reactions for the nonmetals are:

F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V

Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V

Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V

For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.