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After a few hours, the following observations were made:

  • The solution turns green. My guess: $\ce{FeSO4}$ is formed.
  • There are reddish-brown and black powder deposits on both the electrodes.
  • The deposits fall down easily when the set-up is shaken a little.

What are the reactions that are occurring?

According to me, the copper is displaced from the $\ce{CuSO4}$ and is deposited on the iron electrode (the cathode), and $\ce{FeSO4}$ is formed as a green solution. What is getting deposited on the anode? Are there ferrous and ferric oxides/hydroxides being formed?

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  • $\begingroup$ Is there any visible difference between the electrodes, to begin with? $\endgroup$ – Ivan Neretin Jun 14 '18 at 7:09
  • $\begingroup$ @mhchem, there was no visible difference between them, initially. Two normal iron nails were used. They were physically cleaned before conducting the experiment. $\endgroup$ – AcK Jun 15 '18 at 12:15
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The reaction is spontaneous, you don't need put a power supply.

Let's see the reaction with the potentials (obtained from the table of standard reduction potentials):

$$ \begin{align} \ce{Fe^0 &→ Fe^2+ + 2 e-} &\quad E^\circ &= \pu{0.44 V} \\ \ce{Cu^2+ + 2 e- &→ Cu^0} &\quad E^\circ &= \pu{0.337 V} \\ \hline \ce{Fe^0 + Cu^2+ &→ Fe^2+ + Cu^0} &\quad E^\circ &= \pu{0.777 V} \end{align} $$

So, the solid formed in both electrodes is copper. The green solution is related to iron sulfate.

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If you actually passed current thru the cell, you would have been cathodically protecting the cathode - that is, preventing the dissolution of iron. On top of this "cathodic protection", however, you have copper deposition. And the copper could have been loosely deposited as black nano particles.

At the anode, you are discharging SO4-- dianions, which then form H2SO4 (H+ ions plus O2 gas). The acid produced will be corrosive to the iron, so Fe++ will go into solution. You also have the iron metal replacing Cu++ in solution, so some copper may be plating out - similar to the situation at the cathode.

Essentially you have two reactions going on: 1) copper deposition at both electrodes because iron is more active than copper, and 2) more copper deposited at the cathode because of electroplating, and more iron dissolving at the anode for the same reason. More iron dissolving at the anode may not inhibit copper plating out there.

The overall effect at both electrodes can be affected seriously by the current density.

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