Answer to Question #52868 in General Chemistry for tiara
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?
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