0
$\begingroup$

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).

New contributor
Robertsson is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
1
  • $\begingroup$ Your one of many mistakes is posting a question as a snapshot. Please type in your question and make formulating correct using MathJax. $\endgroup$ May 15 at 9:08

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
0

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.