Question

Which species would act as the cathode in the following reaction? ​ 4 Au(s) + 16 Cl-(aq) + 3 O2(g) + 6 H2O(l) → 4 AuCl4-(aq) + 12 OH-(aq)

Select one:
a. Pt(s)
b. Cl-(aq)
c. O2(g)
d. Au(s)

Answer 1

The cathode is d. `"Au(s)"`.

Explanation:

The standard reduction potentials for each half-reaction are:

`ulbb("Standard Reduction Potentials"color(white)(ml)E^@//"V"`
`"AuCl"_4^"-""(aq)" + "3e"^"-" → "Au"(s) + "4Cl"^"-""(aq)" color(white)(m)"+1"`
`"O"_2"(g)" + 2"H"_2"O(l)" + "4e"^"-" → "4OH"^"-""(aq)"color(white)(mll)"+0.4"`

The half-cell with the less positive reduction potential is the anode.

The anode half-reaction is the oxidation of hydroxide ion to oxygen.

We write the cell diagram with the anode on the left.

`"Pt(s)|OH"^"-""(aq)|O"_2"(g)||AuCl"_4^"-""(aq), Cl"^"-""(aq)|Au(s)"`

Thus, the cathode (on the right) is `"Au(s)"`.