enter image description here

Up there is a Electrochemical cell. I have marked the anode (-) and cathode (+) (by taking help from google) but I still need the explanation because such assignment of charges on given electrodes is pretty unacceptable to me.

Here is why.

If the copper electrode is (+) and since we have considered electrons negatively charged, the CU++ ions will not be attracted by copper electrode. Then how come CU++ ions go towards copper electrode and reduction takes place at all? (Like charges repel.)

Actually the book I am reading has not marked the electrodes (+) and (-) by itself however it asks 'will the zinc rod be positive or negative" as I have circled it in the image.

Help taken from following image

enter image description here

  • $\begingroup$ Well, it is clear from your reasoning that the copper electrode can't be (+). Then maybe it can be (-)? $\endgroup$ – Ivan Neretin Jan 24 '18 at 11:03
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    $\begingroup$ Do you really think that people should bend their necks to see your image? $\endgroup$ – Apoorv Potnis Jan 24 '18 at 11:05
  • $\begingroup$ @IvanNeretin I have added an image you can check out dude. $\endgroup$ – Anonymous Jan 24 '18 at 13:28
  • $\begingroup$ I seriously don't see why it got downvoted, it is a brilliant question although a very low level question but the problem is genuine and I am sure many many beginners find and are going to find the same problem and this post is going to help them out. $\endgroup$ – Sufyan Naeem Jan 28 '18 at 10:49

Don't think of the copper rod as positively charged but as less negative.

On both electrodes there's an electrochemical equilibrium: $\ce{M<->M2+ + 2e-}$ (charge may vary depending on the metal)

For more precious metals this equilibrium is further to the left than for non-precious metals. Therefore there are more electrons on the zinc electrode than on the copper electrode which will travel to the less negative electrode in order to compensate for the potential differences.

As ions of more precious metals are more likely to form neutral atoms and there's a constant flow of electrons coming to the copper electrode, the copper gets reduced.

Please note that this answer is strongly simplified.


Anode is the place where the oxidation reaction occur; Cathode is the place where the reduction reaction occur. The reaction is $\ce{Zn +Cu^2+->Cu + Zn^2+}$ so you can tell that Cu is undergoing a reduction reaction, hence it is the cathode.

To elaborate, copper rod's standard potential is more positive than zinc rod's ($E_{\ce{Cu}} = + 0.35\;V, E_{\ce{Zn}} = -0.76\;V$). So, copper will withdraw the electrons from zinc if electrical connection is established. Why? You pointed that out - because of different electrochemical potential.

So a) why is cathode positive and b) why does a reduction of positively charged Cu ions take place at the "positive" cathode?

a) Why is Cu cathode positive?
$\underline{First\; case:\; electrodes\; are\; not\; connected.}$
Immersion of Cu and Zn in the dilute solution of their salts leads to the dissolution of metals and to the formation of an electrolytic double layer (see Galvani potential). Both metal electrodes are negatively charged and the double layer prevents further dissolution of metals because of electrostatic repulsion. In other words - an electrochemical (not chemical!) equilibrium is established. Note that the metal electrodes are both negatively charged and are neither anode nor cathode! However, the chemical potential of the zinc electrode is greater than that of the copper electrode:$$ \mu^*_{Zn}>\mu^*_{Cu}.$$ One can easily see it in the picture where Zn electrode has an excess of electrons. You can think of that as a kind of electron pressure.

${\underline{Second\;case: electrodes\; are\; electrically\; connected.}}$
The two electrodes now have to seek for a new state of equilibrium, with the electrons running from zinc to copper. This happens because the relationship $$ \mu^*_{\ce{Zn}}=\mu^*_{\ce{Cu}}$$ has to be valid. Just imagine that the electrons flow from an area of "high electron pressure" (Zn) into a "low electron pressure" area (Cu). According to the convention, electrons flow from the negative pole to the positive pole. So, copper electrode has a plus sign and the Zn electrode has a minus sign.

b) Why does a reduction of positively charged Cu ions take place at the "positive" cathode?
Once connected, the electrochemical equilibrium (see above) is disturbed at the electrodes. The cells or systems (electrode + solution) try to restore equilibrium. Reaction and equilibrium constant for Cu are: $$ Cu\rightleftharpoons Cu^{2+} + 2e^- $$ $$ K_{Cu}=\frac{\left [\ce{Cu^2+} \right]}{\left[\ce{Cu} \right]} $$ As the concentration of electrons increases, so increases the equlibrium constant. The equilibrium point thus shifts to the left or to the educt side (similar to the Le Chatelier principle). This is the reduction reaction and therefore Cu electrode is the cathode.
Same for Zn: $$ Zn\rightleftharpoons Zn^{2+} + 2e^- $$ $$ K_{Zn}=\frac{\left [\ce{Zn^2+} \right]}{\left[\ce{Zn} \right]} $$ If electron concentration decreases, the equilibrium point shifts to the right or to product side. This is oxidation, and Zn electrode is the anode.

Daniell element

  • $\begingroup$ Maybe let us talk about the potential values of copper and zinc after we get why is copper rod negatively charged? $\endgroup$ – Anonymous Jan 24 '18 at 13:18
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    $\begingroup$ @Mostintelligent If you don't like numbers, focus on the reaction, you do know that the reaction should be $\ce{Zn +Cu^2+=Cu +Zn^2+}$, don't you? $\endgroup$ – Weijun Zhou Jan 24 '18 at 23:55
  • $\begingroup$ what about reaction ? @weijunZhou $\endgroup$ – Anonymous Jan 25 '18 at 13:56
  • $\begingroup$ @ Most intelligent Sorry, it was wrong! Of course, as you said, the cathode has to be positively charged. $\endgroup$ – Viktor Scherf Jan 25 '18 at 15:17
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    $\begingroup$ @Mostintelligent If you know the direction of the reaction (at standard state), you can figure out which side is going through a reduction half reaction and which side is going through an oxidation half reaction, hence figuring out which electrode is a cathode, as explained in this answer. $\endgroup$ – Weijun Zhou Jan 25 '18 at 17:53

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