Say you have two half cells. The electrolytes are connected by a salt bridge and the electrodes by a wire.

  • Half cell 1’s electrolyte is copper(II) nitrate and its electrode is solide copper.
  • Half cell 2’s electrolyte is nitric acid and its electrode is inert (graphite).

I’ve deduced that the reduction reaction will be $$\ce{2 NO3- (aq) + 4 H+(aq) + 2e- -> N2O4(g) + 2 H2O(l)}$$

while the oxidation reaction will be $$\ce{Cu(s)->Cu^2+(aq) + 2e-}$$

Now, the electrode (anode) of half cell 1 is supposed to release electrodes into the external circuit, which end up at the inert cathode in half cell 2. There, nitrate ions will be reduced. Then you’ve got the reordering of ions through the salt bridge etc. etc.

However, I do not understand how such a voltaic cell would be viable. There are nitrate ions already in half cell 1’s electrolyte. Why must electrons flow through the wire then? Can’t the copper electrode directly react with the nitrate ions in the solution it’s submerged in? Can’t the whole redox reaction take place in half cell 1?

  • $\begingroup$ You are overgeneralizing. Yes, some redox reactions can occur in one half-cell; make an electrode out of aluminum or even more active metal and you will see. $\endgroup$ – Ivan Neretin Jun 21 '16 at 4:42
  • $\begingroup$ @IvanNeretin But what about this specific reaction that I described? It was given to me on a test and it got me so confused. $\endgroup$ – lightweaver Jun 21 '16 at 4:51
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    $\begingroup$ This one will not (well, almost). The oxidizing properties of nitrate ion are strongly pH-dependent. $\endgroup$ – Ivan Neretin Jun 21 '16 at 4:54
  • $\begingroup$ Right. Water does not have enough $\ce{H+}$ ions to furnish this particular reaction, but an electrolyte of $\ce{HNO3}$ certainly would. $\endgroup$ – lightweaver Jun 21 '16 at 5:44

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