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Can a galvanic cell with the overall reaction $$\ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)}$$ work?

$\Delta G$ of this reaction is negative (because $E^{\circ}$ of the reaction is $\mathrm{0.34~V}$ as the standard electrode potential of copper will be $\mathrm{-0.34~V}$ as it is acting as an anode). This shows the cell is spontaneous;

However, I have read that $\ce{Cu(s)}$ cannot reduce $\ce{H+}$ so I was wondering if this cell would work?

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TL;DR - What you know from the activity series of metals is still true. You made a mistake in your interpretation of standard electrode potential. The rationale follows:

Standard electrode potentials are written in the reduction direction:

$$\ce{M}^{x+} + x\ce{e- -> M(s)}$$

Standard electrode potentials are measured against the standard hydrogen electrode:

$$\ce{2H+ + 2e- -> H2(g)}$$

Thus for the $\ce{Cu(s)|Cu^{2+}}$ electrode, $E^\circ$ is defined for the followed redox reaction (everything at 1 molar, 1 bar, and in standard states):

$$\ce{Cu^2+ + H2 -> Cu(s) + 2H+}\ \ \ E^\circ = +0.34$$

For the oxidation of copper with acid, the electrode potential would be negative in the direction you wrote it, which translates to a positive $\Delta G^\circ$, since: $$\Delta G^\circ = - nFE^\circ$$

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  • $\begingroup$ wouldn't ∆G be negative since copper usually gets reduced and its E˚ is 0.34. However, when you use it as a anode, the E˚ will change signs, so E˚ anode would become -0.34. When you go to solve E˚ for the cell, it would be E˚cathode - E˚anode. The E˚ for cathode is 0 and E anode is -0.34. Thus, E˚ for the overall cell would be +0.34V (o -- -0.34). Am I doing something wrong in this? $\endgroup$
    – user510
    Commented Feb 16, 2016 at 16:33
  • $\begingroup$ I see your confusion. The equation for $E^\circ_{cell}$ is: $$E^\circ_{cell}=E^\circ_{red,cathode}-E^\circ_{red,anode}$$ The negative sign in the equation accounts for the use of $E^\circ$ for both electrodes in reduction direction. $\endgroup$
    – Ben Norris
    Commented Feb 16, 2016 at 20:15
  • $\begingroup$ So if the anode where aluminum, would the cell work? Aluminums standard electrode potential is -1.66V $\endgroup$
    – user510
    Commented Feb 16, 2016 at 23:06
  • $\begingroup$ Yes. Aluminum would work. $\endgroup$
    – Ben Norris
    Commented Feb 17, 2016 at 23:16
  • $\begingroup$ Just curious... why isn't the negative sign accounted for if you are using, in this case, aluminum as a anode rather than a cathode (which will have negative electrode potential of -1.66V) $\endgroup$
    – user510
    Commented Feb 18, 2016 at 3:39

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