Question

Determine the potential of a concentration cell that is constructed with solutions of Pb(NO3)2 at 0.20 M and 4.0 x 10-4 M

a.0.077 V


b.0.080 V


c.0.084 V


d.0.075 V

Answer 1

The potential of the concentration cell is **c. 0.084 V**.the potential of the concentration cell is 0.084 V**, showcasing a considerable concentration disparity between the solutions, resulting in a higher electrical potential.

Step-by-step explanation

Concentration cells generate electrical potential based on the concentration difference between the two half-cells. In this case, the concentration cell is composed of two solutions of Pb(NO3)2 at differing concentrations: 0.20 M and 4.0 x 10^-4 M. The greater the concentration difference between the two solutions, the higher the potential of the cell. The Nernst equation is applied here to calculate the cell potential:

`\[ Ecell = E°cell - (RT)/(nF) \cdot \ln\left(([Pb^(2+)]_(high))/([Pb^(2+)]_(low))\right) \]`

Given that Pb^2+ ions are common to both solutions, their concentrations
`([Pb^2+]_high and [Pb^2+]_l`ow) are used in the equation.

The Nernst equation in this context becomes:

`\[ Ecell = E°cell - (RT)/(nF) \cdot \ln\left((0.20)/(4.0 * 10^(-4))\right) \]`

Considering standard reduction potentials, the standard cell potential E°cell for Pb(NO3)2 would be 0.13 V. Substituting the values into the equation yields a potential of approximately 0.084 V, indicating that **the potential of the concentration cell is 0.084 V**, showcasing a considerable concentration disparity between the solutions, resulting in a higher electrical potential.