Question

Determine the potential of a concentration cell that is constructed with solutions of pb(no3)2 at 0.20 m and 5.0 x 10-4 m

a. 0.077 v
b. 0.084 v
c. 0.075 v
d. 0.080 v

Answer 1

The potential of a concentration cell can be determined using the Nernst equation. In this case, the potential of the concentration cell with solutions of Pb(NO3)2 at 0.20 M and 5.0 x 10-4 M is 0.084 V.

Step-by-step explanation:

The potential of a concentration cell can be determined using the Nernst equation. In this case, we have a concentration cell constructed with solutions of Pb(NO3)2 at 0.20 M and 5.0 x 10-4 M. The Nernst equation is E = E° - (0.0592/n) * log(Q), where E is the potential, E° is the standard potential, n is the number of electrons transferred, and Q is the reaction quotient. For this concentration cell, both compartments contain the same ion (Pb²+), so the reaction quotient, Q, is equal to the ratio of the concentrations: [Pb²+]/[Pb²+]. Therefore, Q = 5.0 x 10-4 M / 0.20 M = 0.0025. The standard potential for the Pb²+/Pb reduction half-reaction is -0.13 V. Substituting these values into the Nernst equation, we can calculate the potential:

E = -0.13 V - (0.0592/2) * log(0.0025) = -0.13 V - (0.0592/2) * -2.602 = 0.084 V

Therefore, the potential of the concentration cell is 0.084 V.