# How to calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by some concentration?

Consider the cell described below at $$\pu{261 K}$$:

$$\ce{Fe | Fe^2+ (\pu{0.925 M}) || Cd^2+ (\pu{0.981 M}) | Cd}$$

Given $$E^\circ_\ce{Cd^2+ -> Cd} = \pu{-0.403 V}$$, $$E^\circ_\ce{Fe^2+ -> Fe} = \pu{-0.441 V}$$. Calculate the cell potential after the reaction has operated long enough for the $$\ce{Fe^2+}$$ to have changed by $$\pu{0.383 mol L-1}$$.

I have

$$E_\mathrm{cell} = (0.441 - 0.403) - \frac{0.0591}{2}\ln\left(\frac{0.925}{ 0.981}\right) = 0.03974$$

but I'm not sure how to find the $$E_\mathrm{cell}$$ when the $$\ce{Fe2+}$$ concentration has changed by $$\pu{0.383 mol L-1}$$.

• What are the concentrations of the ions at that point? I'm not sure I understand what you mean by "changed by 0.383 mol/L". Is it increasing or decreasing? – Karsten Theis Mar 13 at 1:35
• I added a screenshot of the problem I'm trying to figure out. – LegendOfKass Mar 13 at 1:41
• Are you using the natural or the base-10 logarithm? – Karsten Theis Mar 13 at 3:13
• I'm using the natural log – LegendOfKass Mar 13 at 3:31
• It's either R T / z F ln(Q) or 0.0591 V / z log(Q). For your formula, the term is too large by a factor 2.303 = ln(10). – Karsten Theis Mar 13 at 3:34