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So I failed my test because of this question. It was titled Galvanic Cell although it looks more like an electrolytic cell with the external voltage and everything.

What was asked: explain the function of this cell in detail. some keywords were also presented (not mentioning them all) but there were Gibbs, Faraday, K, and -nFE

I got too confused with the left side while Ni2+ + 2e- -> Ni (s) on the right side looks like a normal reduction in the cathode. What is going on in here?

Standard Reduction Potentials at 25 degrees Celcius:

**Cr2O7- (aq) + 14 H+ (aq) + 6 e- -> 2 Cr3+ (aq) + 7 H20 (l)  +1.33 V
                              Ni2+ (aq) + 2e- --> Ni (s)  -0.25 V**
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  • $\begingroup$ Pt is an inert electrode that will not be involved, what is the potential of the various chromium species? $\endgroup$
    – Burak Ulgut
    Dec 21, 2016 at 11:12
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    $\begingroup$ How about providing some reduction potentials so that we don't have to look them up? $\endgroup$
    – Zhe
    Dec 21, 2016 at 12:53
  • $\begingroup$ EDITED: original post $\endgroup$
    – getbetterinchem
    Jan 13, 2017 at 17:43
  • $\begingroup$ Can you double-check that reduction of dichromate to manganese (II)? $\endgroup$
    – orthocresol ♦
    Jan 14, 2017 at 13:50
  • $\begingroup$ Double-checked. IDK why I had manganese there. Fixed now. $\endgroup$
    – getbetterinchem
    Jan 23, 2017 at 10:56

1 Answer 1

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  • in the case of electrochemical cells/Galvanic cells: Oxidation at the anode (the A and O of electrochemistry, german people will know what I mean), reduction at the cathode (in electrolysis its the other way round)
  • $EMF = E_{red} - E_{ox} = E_{cat} - E_{an} > 0$ for the reaction to run without external voltage (<0 in the case of electrolysis) "the higher redox potential always oxidizes the lower one"

In your case this means running the cell without applying any voltage gives an $EMF = 1.33 - -0.25 V = 1.58 V$

reduction at cathode:

Cr$_2$O$_7^{2-}$$_{(aq)}$ + 14 H$^+$$_{(aq)}$ + 6 e$^-$ $\rightarrow$ 2Cr$^{3+}$$_{(aq)}$ + 7 H$_2$0$_{(l)}$ (in the electrolyte -6e charges)

oxidation at anode:

3 Ni$_s$ $\rightarrow$ 3 Ni$^{2+}_{(aq)}$+ 6e$^-$ (in the electrolyte +6e charges) (already multiplied by 3 for the total reaction)

total:

Cr$_2$O$_7^{2-}$$_{(aq)}$ + 14 H$^+$$_{(aq)}$ + 3 Ni$_s$ $\rightarrow$ 2Cr$^{3+}$$_{(aq)}$ + 7 H$_2$0$_{(l)}$ + 3 Ni$^{2+}_{(aq)}$

Now there is however a clear symbol in the electric cycle standing for a voltage source "V". This is indeed confusing and I would ask what was the meaning of that if the figure was actually titled "Galvanic Cell" which is the other way round. So if it would't be for the title and we would consider running the cell with external voltage, this would then turn around all reactions which I wrote before.

Other things that might be relevant:

  • K$_2$SO$_4$ is responsible for the charge transfer between the cells. Since one cell produces positive charges in solution and the other one negative ones the electrolyte has to establish that the overall charge in both cells stays neutral. The salt bridge connects the electric cycle, without it, the cell won't run for long
  • the Gibb's energy is related to the EMF. $\Delta G = -n\cdot F\cdot EMF$, so again a positive $EMF$ gives a negative $\Delta G$ which means the reaction runs "alone" without external voltage, the Gibb's energy provides the driving force. n is the number of electrons involved in the redox reaction (6) (F is the Faraday constant, a natural constant)
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