If I connect a zinc half cell (Left hand side) and a copper half cell (right hand side), I get an E cell value of 1.1V taken from the voltmeter. If I replace the voltmeter with a light bulb, I will get light and heat energy of the bulb.

I learned that the above set up is equivalent to adding Zinc to copper sulphate solution. However, in this case, there is no wire for the electron to flow...


1.Then, where does the energy go when I add Zinc to copper sulphate solution (that in the first scenario I converted to electrical energy)? Is it released as heat?

2.As the electrons flow from the Zinc half cell to the copper half cell through the light bulb, will the voltage between the electrodes decrease as the charges on the electrodes are supposed to decrease. Then what would the formula for total energy released be?

Please correct me if any of my concepts are wrong. Thanks in advance.

  • $\begingroup$ I might be able to help you with your first question more rigorously, if you are familiar with overpotential, chemical potential and elementary thermodynamics. Are you? $\endgroup$ – Satwik Pasani Jan 3 '14 at 14:32
  • $\begingroup$ @SatwikPasani: I just know a bit on the topics $\endgroup$ – Eliza Jan 3 '14 at 14:40

Ad 1) I am not sure what do you mean by "equivalent". But you add Zinc in a copper sulphate solution, the "energy" will be consumed through cementation.

to a recation Cu$^2$$^+$(aq) + Zn(s) → Cu(s) + Zn$^2$$^+$(aq)

Ad 2) Please see Equilibrium Constant of an Electrochemical Cell Reaction Example Problem where you can find ansewer to your second question.


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.