Question

The $ E{}^\circ $ for half cells $ Fe/F{{e}^{2+}} $ and $ Cu/C{{u}^{2+}} $ are $ -0.44\text{ }V $ and $ +0.32\text{ }V $ respectively. Then:

• A. $ C{{u}^{2+}} $ oxidises $ Fe $
• B. $ C{{u}^{2+}} $ oxidises $ F{{e}^{2+}} $
• C. $ Cu $ oxidises $ F{{e}^{2+}} $
• D. $ Cu $ reduces $ F{{e}^{2+}} $

Answer 1

Answer: (A)

The standard reduction potential of $ F{{e}^{2+}}/Fe, $ is less than that of $ C{{u}^{2+}}/Cu, $ hence Fe is more electropositive than copper. Hence, $ C{{u}^{2+}} $ can oxidise $ Fe $ .