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?

Answer is (e).

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1 Answer 1


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.


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