Calculating Ratio of Stability Constants 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 $\ce{Fe^3+ + e- -&gt; Fe^2+}$ 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&quot;
$$\frac{\beta _1}{\beta _2} = \ce{ \frac{[Fe(CN)_6^{3-}][Fe^2+]}{[Fe(CN)_6^{4-}][Fe^3+]}}$$
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.
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

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

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

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