2
$\begingroup$

Solution I contains a mixture of $\ce{FeCl2}$ and $\ce{FeCl3}$, and solution II contains a mixture of $\ce{K4Fe(CN)6}$ and $\ce{K3Fe(CN)6}$. The concentrations of iron-containing species satisfy the relations $[\ce{Fe^2+}] = [\ce{Fe(CN)_6^4-}]$ and $[\ce{Fe^3+}] = [\ce{Fe(CN)_6^3-}]$. The potential of platinum electrode immersed into the solution I is $0.652\ \mathrm{V}$, while the potential of platinum electrode immersed into solution II is $0.242\ \mathrm{V}$.

Standard redox potential for $\ce{Fe^3+ + e- -> Fe^2+}$ is $0.771\ \mathrm{V}$

Calculate the ratio of the stability constants $\beta [\ce{Fe(CN)_6^3-}] / \beta [\ce{Fe(CN)_6^4-}$]

My attempt

First I wrote out the Nernst Equation for both reactions: $$0.652 = 0.771 - 0.059 \times\mathrm{log}\ce{ \frac{[Fe^2+]}{[Fe^3+]}}$$ $$0.242 = \mathrm{E_{cell}^\circ} - 0.059 \times\mathrm{log}\ce{ \frac{[Fe(CN)_6^{3-}]}{[Fe(CN)_6^{4-}]}}$$

Also I wrote out the expression for the ratio of stability constants" $$\frac{\beta _1}{\beta _2} = \ce{ \frac{[Fe(CN)_6^{3-}][Fe^2+]}{[Fe(CN)_6^{4-}][Fe^3+]}}$$

Now I know that the ratio is just the sum of the 2 logs in the Nernst equations that I have wrote above. However I am not sure how to calculate the $\mathrm{E_{cell}^\circ}$ for the second reaction. Can we assume that $\mathrm{E_{cell}^\circ}$ for the second reaction also equals $0.771\ \mathrm{V}$?

The answer is $8.90 \times 10^6$

Any help is greatly appreciated.

$\endgroup$
4
$\begingroup$

You used all the information from the question except the line that would have helped you solve it

The concentrations of iron-containing species satisfy the relations $[\ce{Fe^2+}] = [\ce{Fe(CN)_6^4-}]$ and $[\ce{Fe^3+}] = [\ce{Fe(CN)_6^3-}]$

From your first cell you find that

$$\frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} = 103.97$$

which means that in your second cell

$$\frac{[\ce{Fe(CN)6^4-}]}{[\ce{Fe(CN)6^3-}]} = 103.97$$

and $E^\circ$ for the cell with cyanide is $+0.361~\mathrm{V}$. (You have a typo in your equation, the 4- should be in the numerator and 3- should be in the denominator. Also your equation is missing units but I won't bother with that.) This is quite far off from the accepted literature value (around $+0.430~\mathrm{V}$) but let's move on.

Now I know that the ratio is just the sum of the 2 logs in the Nernst equations that I have wrote above

No, that is wrong. The ratio of the stability constants refers to that expression you have written, when all of the compounds are placed into the same solution. You cannot simply use the two equations earlier because those describe two separate solutions.

You have to use the difference in $E^\circ$ to work out the ratio of $\beta$.

The way to do this would be to realise that if you take the difference between the $E^\circ$ you get

$$\begin{align} \ce{Fe^3+ + Fe(CN)6^4-} &\longrightarrow \ce{Fe^2+ + Fe(CN)6^3-} & E^\circ &= +0.410~\mathrm{V} \end{align}$$

The equilibrium constant for this reaction is the quantity that you are looking for

$$K = \frac{[\ce{Fe^2+}][\ce{Fe(CN)6^3-}]}{[\ce{Fe^3+}][\ce{Fe(CN)6^4-}]}$$

and so

$$K = \exp\left(\frac{-\Delta_\mathrm{r} G^\circ}{RT}\right) = \exp\left(\frac{-nFE^\circ}{RT}\right) = 8.589 \times 10^6$$

$\endgroup$
  • $\begingroup$ Thanks for the great answer! You say that I can't use the two Nernst equations because those describe two separate solutions. I am curious to know if these two solutions were part of a galvanic cell, where one of them is the cathode and other is the anode. Would I able to use the Nernst equations now since the two solutions are actually in equilibrium with each other. $\endgroup$ – Nanoputian Jun 10 '16 at 7:26
  • $\begingroup$ It would have to be in equilibrium for that to happen (i.e. you have to let current flow until there is no more current). Galvanic cell systems reach equilibrium slowly because the current flows slowly, which is what allows the potential difference to be determined. Think about the Nernst equation - because the reaction proceeds very slowly, Q tends towards K very slowly, which makes E very stable. You would have to allow electricity to flow until Q = K before you could use the Nernst equation to determine K. $\endgroup$ – orthocresol Jun 10 '16 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.