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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 $\pu{0.383 mol L-1}$.

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    $\begingroup$ 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? $\endgroup$ – Karsten Theis Mar 13 at 1:35
  • $\begingroup$ I added a screenshot of the problem I'm trying to figure out. $\endgroup$ – LegendOfKass Mar 13 at 1:41
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    $\begingroup$ Are you using the natural or the base-10 logarithm? $\endgroup$ – Karsten Theis Mar 13 at 3:13
  • $\begingroup$ I'm using the natural log $\endgroup$ – LegendOfKass Mar 13 at 3:31
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    $\begingroup$ 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). $\endgroup$ – Karsten Theis Mar 13 at 3:34

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