0
$\begingroup$

We just had a lab experiment on galvanic cells and we used 0.5 M and 1 M iron(III) nitrate solution at the anode (iron electrode) and 1 M and 0.5 M copper(II) nitrate solution at the cathode (copper electrode) for runs 1 and 2, respectively.

How can I determine which redox reaction occurred? There are many possible reactions $(\ce{Cu^2+ + e- -> Cu+}$ or $\ce{Cu^2+ + 2e- -> Cu}).$ The experimental cell potential we got was 0.48 V and 0.44 V. Temperature that we recorded was 28 °C and 29 °C for runs 1 and 2.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Here, Fe(III) will be reduced to Fe(II), and Copper(0) will be oxidized to Cu(II). The standard potential for Cu2+/Cu is +0.34 V, and for Fe3+/Fe2+ it is +0.77 V. Nernst's law gives the potentiels for a 0.5 M solution. It is +0.34 - 0.009 V = 0.33 V pour the Copper electrode, and 0.77 - 0.018 = 0.75 V for the Fe3+/Fe2+ electrode. With this choice, you get nearly your results. It you had chosen another system of half-reactions, you will get results which are far away from your measurements.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Question, why did the Fe undergo reduction despite it being in the anode? I thought that in galvanic cells, anodes undergo reduction? $\endgroup$
    – Polo Jerome Daquipil
    Commented Oct 24, 2019 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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