Question

. At STP, a galvanic cell was set up having the following half-reactions.
Fe2+(aq) + 2e- → Fe(s) -0.41
Cu2+(aq) + 2e-→ Cu(s) 0.34

The copper half-cell contained 100.0 mL of 1.00 M CuSO4, The lead half-cell contained 50.0 mL of 0.100 M FeSO4, to which was added 50.0 mL of 0.300 M NaOH. The cell potential was measured to be 1.155 V. What is the concentration of Fe2+ in the iron half-cell?

Answer 1

Answer on Question #52868 – Chemistry – General chemistry

Question:

At STP, a galvanic cell was set up having the following half-reactions.

$$\mathrm{Fe}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}) -0.41$$ $$\mathrm{Cu}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) 0.34$$

The copper half-cell contained $100.0\,\mathrm{mL}$ of $1.00\,\mathrm{M}\,\mathrm{CuSO}_4$, The lead half-cell contained $50.0\,\mathrm{mL}$ of $0.100\,\mathrm{M}\,\mathrm{FeSO}_4$, to which was added $50.0\,\mathrm{mL}$ of $0.300\,\mathrm{M}\,\mathrm{NaOH}$. The cell potential was measured to be $1.155\,\mathrm{V}$. What is the concentration of $\mathrm{Fe}^{2+}$ in the iron half-cell?

Solution:

$$E = E^0 - \frac{RT}{nF} \log \frac{c(\mathrm{Fe}^{2+})}{c(\mathrm{Cu}^{2+})}$$ $$E = E^0 - \frac{0.0592}{2} \log \frac{c(\mathrm{Fe}^{2+})}{c(\mathrm{Cu}^{2+})}$$ $$E^0 = \varphi(\mathrm{Cu}^{2+}/\mathrm{Cu}) - \varphi(\mathrm{Fe}^{2+}/\mathrm{Fe}) = 0.34 + 0.41 = 0.75$$ $$1.115 = 0.75 - \frac{0.0592}{2} \log \frac{c(\mathrm{Fe}^{2+})}{c(\mathrm{Cu}^{2+})}$$ $$\log \frac{c(\mathrm{Fe}^{2+})}{c(\mathrm{Cu}^{2+})} = -2 \cdot \left( \frac{1.155 - 0.75}{0.0592} \right)$$ $$\frac{c(\mathrm{Fe}^{2+})}{c(\mathrm{Cu}^{2+})} = 2 \cdot 10^{-14}$$ $$c(\mathrm{Fe}^{2+}) = 2 \cdot 10^{-14} \,\mathrm{mol}\,\mathrm{L}^{-1}$$

Such a low concentration can be explained with the formation of insoluble iron (II) hydroxide with NaOH addition:

$$\mathrm{Fe}^{2+} + 2\mathrm{OH}^{-} = \mathrm{Fe}(\mathrm{OH})_2 \downarrow, \quad K_s = 1.4 \cdot 10^{-15}$$

Answer: $2 \cdot 10^{-14}\,\mathrm{mol}\,\mathrm{L}^{-1}$

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