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I am reading up on how galvanic cell works and I realised that the flow of electrons is from Zn to Cu. But Zn is more electronegative compared to Cu according to periodic table trends

I read it somewhere that when comparing transition metals, it may be different, but it was not further elaborated. Can someone explain this to me how Cu is more electronegative than Zn?

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2 Answers 2

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There are two common confusions in your question. I hope I can clear them up.

Electrons are not flowing from zinc to copper. Not exactly.

In the Zinc-Copper galvanic cell, the net reaction is:

$$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}$$

Zinc is being oxidized:

$$\ce{Zn(s)->Zn}^{2+}(aq) \ce{+2e-} $$

Copper is being reduced:

$$\ce{Cu}^{2+}(aq)\ce{+2e- ->Cu(s)}$$

Electrons, therefore are being transferred from zinc not to copper, but copper (II) ions, which are different than the copper atoms in Cu(s). We need to be comparing something about zinc atoms and copper (II) ions.

Electronegativity has nothing to do with electrochemistry.

Electronegativity is the attraction an atom has for a pair of electrons in a covalent bond.

Electrochemistry is about controlling the electron transfer that accompanies a redox reaction. Usually this involves ionic species, and so electronegativity is not the relevant quantity to compare. In this case, we care about the energy change that occurs when copper and zinc lose electrons (ionization energy) and the stability of the resulting ions in solution. Ionization data mined from this NIST publication.

For zinc: $$\ce{Zn->Zn+ + e-} \ \ \ \ \ \Delta H= 9.585\times 10^2 \text{ kJ/mol}$$ $$\ce{Zn+->Zn}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.733\times 10^3 \text{ kJ/mol}$$ $$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{ kJ/mol}$$

For copper: $$\ce{Cu->Cu+ + e-} \ \ \ \ \ \Delta H= 7.454\times 10^2 \text{ kJ/mol}$$ $$\ce{Cu+->Cu}^{2+}\ce{ + e-} \ \ \ \ \ \Delta H= 1.958\times 10^3 \text{ kJ/mol}$$ $$\ce{Cu->Cu}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.703\times 10^3 \text{ kJ/mol}$$

Why is zinc oxidized and copper reduced?

Because doing so leads to a lower energy state. It takes more energy to remove two electrons from copper than to remove two electrons from zinc. Thus, to get a negative energy change (spontaneous), when need to add the zinc reaction as written and the copper reaction in reverse:

$$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ \ \ \Delta H= 2.692\times 10^3 \text{ kJ/mol}$$ $$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ \ \ \Delta H= -2.703\times 10^3 \text{ kJ/mol}$$ $$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ \ \Delta H= -12\times 10^3 \text{ kJ/mol}$$

The estimate above assumes that the solvation energies of both ions are similar and their standard entropies are similar. I could not find a good source of those data.

In electrochemistry, we care more about the potential of the cell, $E_{cell}$, which is the potential for the cell to do electrical work. Potential, $E$, is related to free energy change $\Delta G$ by the following equation, where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant $F=96,485 \frac{\text{J}}{\text{V mol}}$.

$$\Delta G=-nFE$$

For a process to be spontaneous, $\Delta G$ must be negative. $\Delta G$ is negative when $E$ is positive. Thus, in a galvanic cell, electrons will flow in the direction that produces a positive cell potential.

How can we determine which direction that is? First, we could set up the zinc-copper cell, and connect it into a circuit and let it run. In this cell, a piece of copper metal serves as one electrode and a piece of zinc metal serves as the other. Whichever electrode increased in mass is the electrode at which reduction occurred: metals cations are converted to metal atoms, coming out of solution and increasing the mass of the electrode. The other electrode must be the one responsible for oxidation (metal atoms become cations and dissolve away); the mass of this electrode must decrease.

Or, we could do a simpler experiment. Dip a piece of copper metal into a solution of Zn2+. Nothing exciting happens (no reaction). Then, dip a piece of zinc metal into a solution of Cu2+. Zinc reduces the Cu2+ ions and copper metal plates onto the zinc.

Or, we could rely on tables of standard reduction potentials for half cells, measured against the standard hydrogen electrode SHE (for which E is defined to be ZERO). Standard reduction potentials come from Wikipedia, which mines them from a number of legitimate sources.

SHE: $$\ce{H2(1 atm) + 2e- -> 2H+(1 M)} \ \ \ E^o =0 \text{ V}$$. For zinc: $$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=-0.762 \text{ V}$$ For copper: $$\ce{Cu->Cu}^{2+} \ce{+ 2e-} \ \ \ E^o =+0.34 \text{ V}$$ To generate a spontaneous reaction, these half cells are combined in such a way so that the overall cell potential is positive: $$\ce{Zn->Zn}^{2+} \ce{+ 2e-} \ \ \ E^o=+0.762 \text{ V}+$$ $$\ce{Cu}^{2+} \ce{+ 2e- -> Cu(s)} \ \ \ E^o=+0.34 \text{ V}$$ $$\ce{Zn(s) + Cu}^{2+} (aq) \ce{-> Zn}^{2+} (aq)\ce{ + Cu(s)}\ \ E^o=+1.10 \text{ V}$$ This analysis only applies to standard conditions. Under nonstandard conditions (for example [Zn2+]>>[Cu2+], then the reverse reaction could be spontaneous.

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The answer by Ben Norris is nice but it seems to have an additional confusion. The difference between reduction potentials of copper and zinc can be calculated based on the empirical ionization data that Ben describes, but this doesn't mean that the difference has nothing to do the electronegativity of the two different elements. A nice discussion of this issue is provided at the following link: http://blogs.scientificamerican.com/degrees-of-freedom/2011/10/13/the-periodic-table-and-batteries/

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Could you try to elaborate on this; maybe paraphrasing/summarizing the link here? –  ManishEarth Jan 1 at 18:04

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