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Say the following reactions are occurring in my voltaic cell:

$$\ce{Mg->Mg^{2+} + 2e- +2.36V}$$ $$\ce{Cu^{2+} +2e-->Cu -0.34V}$$

Then, the Nernst equation would become:

$$E_{cell}=2.70-\frac{RT}{nF}ln\frac{[Mg^{2+}]}{[Cu^{2+}]}$$

However, if the initial volumes and concentrations of the copper and magnesium solutions were equal, the fraction within the natural logarithm would constantly equal 1.

Does this mean there is no change in voltage as the concentrations decrease, or did I understand something wrong?

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  • $\begingroup$ You got it wrong. As the reaction progresses, [Mg2+] changes; so does [Cu2+], and also their ratio, and also ln thereof. $\endgroup$ – Ivan Neretin Nov 27 '18 at 6:07
  • $\begingroup$ @IvanNeretin thank you, I just realised I was thinking of the Mg2+ and Cu ratio $\endgroup$ – George Tian Nov 27 '18 at 6:59

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