Suppose you have a cell set up between a copper metal/copper(II) ion electrode and a reference electrode.

Under standard conditions, the emf of this cell was −0.07 V. The standard electrode potential of the copper metal / copper(II) ion electrode is +0.34 V. Hence the standard electrode potential of the reference electrode is:

A: -0.41 V

B: -0.27 V

C: +0.27 V

D: +0.41 V

My attempt:

I was taught this equation: $\mathscr{E} _{cell} = \mathscr{E} _{a} - \mathscr{E} _{b}$


$\mathscr{E} _{a}$ is the emf of the more positive electrode potential

$\mathscr{E} _{b}$ is the emf of the less positive electrode potential

But using this equation, none of the combinations of electrode potentials give -0.07V as the emf.



1 Answer 1


Chapter 14 of Harris' Quantitative Chemical Analysis has a helpful section entitled, an intuitive way to think about cell potentials. Here's the corresponding graphic:

enter image description here

In this case, the author is describing a Galvanic cell with cadmium as the anode and silver as the cathode. The potentials of the two half-cells are known and marked on the line-graph. Electrons always flow to the more positive electrode, which in this figure is towards the right. The difference between the two half-cells is the cell potential.

In your case, you know the separation (-0.07 V) and one of two points (+0.34 V). If we ignore the sign for the moment, we can use the line-graph to narrow the possibilities down to C or D, since each of these differ from +0.34 V by 0.07. The question now is, which direction are the electrons flowing? When determining the cell potential using this line graph, we subtract the right-most number from the left-most number in order to obtain a positive potential. Since the cell potential in your case is negative, we must be doing the opposite. Therefore, the reference electrode potential is +0.41 V (D).

  • $\begingroup$ I understand it slightly better now, thanks, but I'm still unsure on why we do the opposite for a negative potential. Is it because the electrons are flowing to the less positive electrode in my case instead of to the more positive one so the calculation is the opposite? $\endgroup$
    – Cobbles
    May 26, 2015 at 10:52
  • $\begingroup$ @Cobbles it is a "frame of reference". If you are 2 meters away from something, you may say that it is 2 m in front of you. If you turn around, it is 2 m behind you. The object hasn't moved, just your perception of "in front" and "behind". We measure potentials versus a reference electrode, so regardless of where the reference electrode potential lies, we determine the cell potential as working minus reference. $\endgroup$ May 26, 2015 at 11:04

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