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Calculating the EMF of batteries

  1. Aug 18, 2011 #1
    Hi all,

    I'm a total noob when it comes to chemistry, and would need your help to understand the calculation of EMF for batteries.

    I have a couple of questions about the calculation for Zinc-Carbon batteries in Wikipedia:

    I can't manage to get 1.5V as a result...

    So, here's a couple questions:

    1) Anode reaction: Zn(s) → Zn2+(aq) + 2 e- [e° = -1.04 volts]
    Why -1.04 ? Why not +0.76 like in all the textbooks and other webpages?
    Like on this page: http://www.ausetute.com.au/calcelemf.html

    2) Cathode reaction:
    2MnO2(s) + 2 e- + 2NH4Cl(aq) → Mn2O3(s) + 2NH3(aq) + H2O(aq) + 2 Cl- [e° ≈ +.5 v]
    How to get the 0.5V?

    Thanks already for your help!
  2. jcsd
  3. Aug 19, 2011 #2


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    Staff: Mentor

    -0.76 (not +0.76) is a standard potential - that means it will be observed when activity of Zn2+ is exactly 1. As a first approximation that means concentration of Zn2+ equal to 1M - if the concentration is different, observed potential will be different (it can be calculated from the http://en.wikipedia.org/wiki/Nernst_equation).

    Same for the cathode - to calculate potential you need to take into account activities of all substances present. In concentrated solutions (and that's what you find in the battery, it is filled with NH4Cl paste) these are in practice impossible to calculate and it is easier to measure.
  4. Aug 22, 2011 #3
    Thanks a lot! Very good explanation!
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