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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.

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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.

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    $\begingroup$ 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 Oct 24 '19 at 10:55

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