Question

describe what is happening at the anode and cathode in a voltaic cell between copper (submerged in copper (ii) sulfate) and iron (submerged in iron (ii) sulfate). reference appendix l in the openstax textbook.

Answer 1

In a voltaic cell with copper and iron electrodes in sulfate solutions, iron is oxidized at the anode, losing electrons, while copper(II) ions are reduced to copper metal at the cathode by gaining electrons.

Step-by-step explanation:

Understanding the Voltaic Cell Operations at Anode and Cathode

In a voltaic cell, such as the one between copper and iron submerged in their respective sulfate solutions, we witness an electrochemical reaction where electrons are transferred between the two different metals. These electrons flow through an external circuit from the anode to the cathode.

At the anode, which in this case would be the iron electrode, oxidation occurs. Iron atoms lose electrons to form iron(II) ions (Fe2+), thus releasing electrons into the external circuit. The half-reaction that occurs at the iron anode can be represented as:
Fe(s) → Fe2+(aq) + 2e−

On the other hand, at the cathode, the copper(II) ions (Cu2+) from the copper(II) sulfate solution gain electrons and are reduced to solid copper. This reduction half-reaction takes place on the surface of the copper electrode and can be written as:
Cu2+(aq) + 2e− → Cu(s)

In summary, iron is oxidized at the anode, losing electrons, which travel through the external circuit to the cathode, where copper ions are reduced, gaining electrons to become solid copper.