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If the concentration of sn2+ in the cathode compartment is 1.50 m and the cell generates an emf of 0.24 v , what is the concentration of pb2+ in the anode compartment?

Respuesta :

E° cell = E° cathode - E° anode
=-0.14 +0.13
= -0.01V
Q= [pb⁺²]/[Sn⁺²] = Q= [pb⁺²]/1.50]
E = E° = -0.05916/n log Q
0.24 = -0.01 = 0.05916/2 log Q
Log Q = -8.452
Q = +3.53 × 10⁻⁹
(pb +2) = 3.53 × 10⁻⁹ × 1.5
=5.297 × 10⁻⁹M
[pb⁺²] = 5.3 × 10⁻⁹M
10freakygirls

Answer:

4.8*10^-8M

Explanation:

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