# Determining an unknown E° from two similar half ox/redox reactions

Use the oxidation half-cell and reduction half-cell, $$\ce{Cr^3+/Cr^2+} = \pu{0.424 V}$$ and $$\ce{Cr^3+/Cr2O7^2-} = \pu{1.32 V}$$, to determine the $$E^\circ$$ for $$\ce{Cr2O7^2-/Cr^2+}$$:
a. $$\pu{-1.75 V}$$
b. $$\pu{-0.02 V}$$
c. $$\pu{0.81 V}$$
d. $$\pu{1.75 V}$$
e. $$\pu{-1.10 V}$$

I will have to obtain the $$E_\mathrm{cell}^\circ$$ for $$\ce{Cr2O7^2-/Cr^2+}$$ only from these two half reactions, $$\ce{Cr^3+/Cr^2+} = \pu{0.424 V}$$ and $$\ce{Cr^3+/Cr2O7^2-} = \pu{1.32 V}$$.

This problem is similar to one of the practice examples in the book:

In an acidic solution, $$\ce{O2(g)}$$ oxidizes $$\ce{Cr^2+(aq)}$$ to $$\ce{Cr^3+(aq)}$$.
The $$\ce{O2(g)}$$ is reduced to $$\ce{H2O(l)}$$.
The $$E^\circ_\mathrm{cell}$$ for the reaction is $$\pu{1.653 V}$$.
What is the standard electrode potential for the couple $$\ce{Cr^3+/Cr^2+}$$?

But in this practice example the solution is to simply withdraw the total cell potential from one of the half reactions.

This is clearly not the case here since I do not get the right answer from any plus or minus actions.

So I figured I'd maybe have to use one of these equations:

$$\ce{Cr2O7^2- + 14 H+ + 6e- -> 2Cr^3+ + 7 H2O }$$ $$\ce{Cr^3+ + e- -> Cr^2+}$$

And maybe multiply/divide one of the voltages with the number of electrons in said reaction.

Would that be the correct approach?

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• Your one of many mistakes is posting a question as a snapshot. Please type in your question and make formulating correct using MathJax. May 15 at 9:08

First the equation of the reduction of $$\ce{Cr2O7^{2-}}$$ into $$\ce{Cr^{3+}}$$ does not produce basic ions $$\ce{OH-}$$ as given in Robertson's text. The reason is that, if such basic ions were produced, they would immediately react with $$\ce{Cr^{3+}}$$ to produce a precipitate of $$\ce{Cr(OH)3}$$. So the equation would not produce the ion $$\ce{Cr^{3+}}$$ but the precipitate $$\ce{Cr(OH)3}$$. To prevent this precipitation, the reduction of dichromate ion will proceed in acidic solution according to :

$$\ce{Cr2O7^{2-} + 14 H+ + 6 e- -> 2 Cr^{3+} + 7 H2O}$$

The redox potential relative to this equation is : $$\ce{E° = +1.33}$$ V. The corresponding value of Delta G° is : $$\Delta \mathrm{G}°_1 =\pu{- z·E·F = - 6 *1.33 V *96500 Cb = -770 100 J/mol}$$.

The wanted final equation is the sum of the two equations : $$\ce{Cr2O7^{2-} + 14 H+ + 6 e- -> 2 Cr^{3+} + 7 H2O}$$ $$\ce{2 Cr^{3+} + 2 e- -> 2 Cr^{2+}}$$ The redox potential of the $$\ce{Cr^{3+}/Cr^{2+}}$$ equation is $$-0.424$$ V. The corresponding value of $$\Delta \mathrm{G}°_2 = \pu{- 2 ({-0.424} V) {96500} Cb}$$ = $$81 800$$ J/mol.

The wanted final equation is then : $$\ce{Cr2O7^{2-} + 14 H+ + 8 e- -> 2 Cr^{2+} + 7 H2O}$$

The total and final $$\Delta \mathrm{G}°_f$$ is the sum of $$\Delta \mathrm{G}°_1$$, and $$\Delta \mathrm{G}°_2$$ :

$$\Delta \mathrm{G}°_f = -770100 + 81800 = -688 300$$ J/mol.

The corresponding final redox potential is :

$$\ce{E°_f = - \Delta \mathrm{G}°_f /8F = \frac{688 300 J/mol}{8·96500 Cb} = 0.8915 V}$$

There must be a mistake somewhere, because this final value does not correspond to any of the five proposed solutions.