K(s) + 1/2 Cl2(g) → KCl(s) ΔH = -437 kJ/mol The elements K and Cl react directly to form the compound KCl according to the equation above. Refer to the information above and the table to answer the questions that follow.

Cl2(g) + 2e- → 2Cl-(g)

Which of the following expressions is equivalent to ΔHo for the reaction represented above?

Process ΔH° (kj/mol)
K(s) → K(g) v
K(g) → K+(g) + e^- w
Cl2(g) → 2Cl(g) x
Cl(g)+ e^- → Cl^-(g) y
K+(g) + Cl(g)^- → KCL z

the mass of the cathode increases because ions from the solution get reduced to atoms and become part of the cathode.

explanation:

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a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

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