Asked by bear
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
Pb^2+ e= -0.13 volts M=?
Ni^2+ e=-.25 volts Ni^2 M=.27M
e=-0.020 volts
Answers
Answered by
DrBob222
You have too many error in your post. Electrons not right, charges not right.
Answered by
bear
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
Answered by
DrBob222
See my response to your repost above. Hope that helps.
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