Question

Determine the concentration (in M ) of lead ions

Pb^2+ e= -0.13 volts M=?

Ni^2+ e=-.25 volts Ni^2 M=.27M

e=-0.020 volts

Answer 1

You have too many error in your post. Electrons not right, charges not right.

Answer 2

Pb(s) Pb^2+ (aq) [pb^2+] =?M

Ni^2+(aq) Ni^2+=.27M Ni(s)
Pb^2+ +2e^- Pb E= -0.13v
Ni^2+ +2e^- Ni E=-0.25

E=-0.020v

Answer 3

See my response to your repost above. Hope that helps.

Answer 4

[AI-generated answer — Explain Bot]

To determine the concentration of lead ions (Pb^2+) in Molarity (M), we can use the Nernst equation, which relates the standard cell potential (E°), the actual cell potential (E), the concentration of the ion (M), and the number of electrons transferred in the reaction (n):

E = E° - (0.0592 V/n) * log(M)

Given:
E = -0.13 volts (for Pb^2+)
E° = -0.020 volts (for unknown ion)
n = 2 (because 2 electrons are transferred for Pb^2+)

We need to rearrange the equation to solve for the concentration (M):

M = 10^((E° - E) / (0.0592 V/n))

Substituting the given values:

M = 10^((-0.020 V - (-0.13 V)) / (0.0592 V/2))

M = 10^(-0.11 V / (0.0296 V))

M = 10^(-3.7162)

M ≈ 2.87 x 10^(-4) M

Therefore, the concentration of lead ions (Pb^2+) is approximately 2.87 x 10^(-4) M.