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
3 answers
You have too many error in your post. Electrons not right, charges not right.
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
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
See my response to your repost above. Hope that helps.