3
$\begingroup$

Let's take a simple Daniell cell for example: Daniell cell

Where does the energy required to drive electrons through the external circuit and power up devices (like a light bulb) come from? Does the oxidation of zinc release that energy? And, does the reduction of copper result in more energy being released? (since we're calculating the cell potential by adding up the oxidation and reduction potentials of the two half cells). I've always wondered about this question, and I'd really appreciate a clear and concise explanation regarding this particular topic.

$\endgroup$
1
$\begingroup$

Judging by your above comment to beginner's answer, you seem to be confused about how reactions proceed in general.

Take a look at this graph : http://en.wikipedia.org/wiki/Activation_energy#/media/File:Activation_energy.svg

The oxidation in the beginning certainly requires energy, but at the end of the reaction, the system ends up at a lower energy state overall. This causes the reaction to be "spontaneous".

$\endgroup$
0
$\begingroup$

The cell is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. In other words, the metal atoms of one half-cell are able to induce reduction of the metal cations of the other half-cell; conversely stated, the metal cations of one half-cell are able to oxidize the metal atoms of the other half-cell. When metal B has a greater electronegativity than metal A, then metal B tends to steal electrons from metal A (that is, metal B tends to oxidize metal A), thus favoring one direction of the respective reaction.

$\endgroup$
  • $\begingroup$ I appreciate your answer. Yet that doesn't fully answer my question. What I'd like to know is: Wouldn't it require energy to expel an electron from an atom rather than releasing it? So, how exactly does oxidizing an element produce energy? $\endgroup$ – Heba Apr 13 '15 at 18:25
  • $\begingroup$ I recommend that the new last question post it in an separate post and try to clarify your doubt a little further. Because, as your formulated question is still quiet broad. However, I though that the shared question in this post is already answered. Hope you agree and have a good day :D $\endgroup$ – Another.Chemist Apr 16 '15 at 4:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.