Let's break down and solve each part step by step. --- ## **Given:** - **One end of iron wire in .02 M Fe²⁺ solution.** - **Other end in .005 M Fe²⁺ solution.** - **Both solutions are separated by a porous wall.** The relevant half-cell reaction: \[ \text{Fe}^{2+} + 2e^- \rightleftharpoons \text{Fe}(s) \] The standard electrode potential for this half-cell: \[ E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -.44\, \text{V} \] --- ## **(a) Which end of the wire will corrode?** **Corrosion** occurs at the **anode** (oxidation site). At the anode: \[ \text{Fe}(s) \rightarrow \text{Fe}^{2+} + 2e^- \] We have two half-cells: - **Cell 1:** Fe in .02 M Fe²⁺ - **Cell 2:** Fe in .005 M Fe²⁺ The end in the **lower concentration** of Fe²⁺ will have a **lower reduction potential** and will act as the **anode** (where oxidation and thus corrosion occurs). **Therefore:** - The end in **.005 M Fe²⁺** will corrode. --- ## **(b) What will be the potential difference between the two ends?** Use the **Nernst equation** for each half-cell: \[ E = E^\circ + \frac{.059}{n} \log [\text{Fe}^{2+}] \] where \( n = 2 \). **For each end:** - **End 1:** \([Fe^{2+}] = .02\, M\) - **End 2:** \([Fe^{2+}] = .005\, M\) ### **Potential at End 1:** \[ E_1 = -.44 + \frac{.059}{2} \log(.02) \] \[ \log(.02) = \log(2 \times 10^{-2}) = \log 2 + \log 10^{-2} = .301 - 2 = -1.699 \] \[ E_1 = -.44 + .0295 \times (-1.699) \] \[ E_1 = -.44 - .0502 = -.4902 \, \text{V} \] ### **Potential at End 2:** \[ E_2 = -.44 + \frac{.059}{2} \log(.005) \] \[ \log(.005) = \log(5 \times 10^{-3}) = \log 5 + \log 10^{-3} = .699 - 3 = -2.301 \] \[ E_2 = -.44 + .0295 \times (-2.301) \] \[ E_2 = -.44 - .0679 = -.5079 \, \text{V} \] ### **Potential Difference** \[ \Delta E = E_1 - E_2 = (-.4902) - (-.5079) = .0177 \, \text{V} \] --- ## **Summary Table** | End of Wire | [Fe²⁺] (M) | Electrode Potential (V) | Corrosion? | |:-----------:|:----------:|:----------------------:|:-----------:| | 1 | .02 | -.4902 | No | | 2 | .005 | -.5079 | **Yes** | --- ## **Final Answers** ### **(a)** - **The end of the wire in .005 M Fe²⁺ will corrode.** ### **(b)** - **The potential difference between the two ends is .018 V (rounded to two significant digits), with the .02 M end positive relative to the .005 M end.**