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This reaction proceeds in the galvanic cell:

$$\ce{Zn + Pb^{2+} -> Zn^{2+} + Pb}$$

I need the equation for the half-reactions of this cell and the theoretical voltage produced by this cell if concentration of $\ce{Zn^{2+}}$ is $\mathrm{0.1\ mol/L}$ and concentration of $\ce{Zn^{2+}}$ is $\mathrm{10^{-4}\ mol/L}$.

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closed as off-topic by Klaus-Dieter Warzecha, Geoff Hutchison, Philipp, ron, jerepierre Dec 16 '14 at 16:02

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$$Pb^{2+} + 2e^{-}->Pb E_{0}(Pb^{2+}) = -0.1263$$ $$Zn->Zn^{2+} +2e^{-} E_{0}(Zn^{2+}) = -0.76 $$

$$[Zn^{2+}] = 0.1 mol/L [Pb^{2+}] = 10^{-4} mol/L$$

$$Qc= \frac{[Zn^{2+}]}{[Pb^{2+}]}$$ $$E_{0}(cell) = E_{0}(Pb^{2+})- E_{0}(Zn^{2+})= -0.1263 - (-0.76) $$
$$ E(cell)= E_{0}(cell) - \frac {RT ln Q_{c}}{nF}$$

Small piece of advice : Tag homework and exercises on these type of questions , read through "Help Center" specifically "Asking" . Try to attempt the question by yourself and show your own attempt of an answer. This site isnt "Solve my homework site " good luck cheers

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  • $\begingroup$ Thank you, I tried and submit, teacher said it is wrong ,thats why i wrote it here. $\endgroup$ – Samber23 Dec 16 '14 at 14:03
  • $\begingroup$ Well if you did something post that here , only then people will try to answer i think . anyway this is still in beta so i guess moderators arent here yet .or your post would have been downvoted/removed $\endgroup$ – Gowtham Dec 16 '14 at 14:08

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