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Let's solve this **step by step**. The cell is: [ Fe, Fe^{2+} (0.1M) ;|; Cd^{2+} (0.001M), Cd ] ### Step (i): Write the cell reaction We have: * **Fe/Fe²⁺** electrode * **Cd/Cd²⁺** electrode Standard reduction potentials: [ E^\circ_{Cd^{2+}/Cd} = -0.40 \text{ V}, \quad E^\circ_{Fe^{2+}/Fe} = -0.44 \text{ V} ] The **more negative reduction potential** will oxidize (anode), the less negative will _____ ________ ______ _____ _____ __________ __________.
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