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Bad White [126]
3 years ago
10

Use the standard reduction potentials located in the Tables' linked above to calculate the equilibrium constant for the reaction

:
Cd^2+(aq) + H2(g) ----------->Cd(s) + 2H+(aq)
Chemistry
1 answer:
Nadya [2.5K]3 years ago
4 0

Answer:

3.1 * 10^-14

Explanation:

Note that E°cell = 0.0592/n log K

We can obtain E°cell from the standard reduction potentials of cadmium and hydrogen

Anode reaction

H2(g) ----> 2H+ + 2e

Cathode reaction

Cd^2+(aq) + 2e -----> Cd(s)

E°cell = E°cathode - E°anode

E°cathode = –0.40 V

E°anode = 0 V

E°cell =  –0.40 V - 0 V

E°cell = –0.40 V

E°cell = 0.0592/n log K

Where n=2 electrons transferred

–0.40 = 0.0592/2 log K

–0.40 = 0.0296 log K

log K = –0.40/0.0296

log K = -13.5135

K = Antilog ( -13.5135)

K = 3.1 * 10^-14

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