• ↓ E° → oxidation → anode
• ↑ E° → reduction → cathode
Fe(NO3)2 → Fe2+ + 2 NO3-
Fe2+(aq) + 2 e- → Fe(s) E° = –0.447 V → anode
Find the half-reaction with higher E° value than Fe2+/Fe:
CuCl2 → Cu2+ + 2 Cl-
Cu2+(aq) + 2 e- → Cu(s) E° = 0.3419 V
NiCl2 → Ni2+ + 2 Cl-
There are three voltaic cells. In each voltaic cell one half-cell contains a 1.0 M Fe(NO3)2(aq) solution with an Fe electrode. The contents of the other half-cells are as follows:
In which voltaic cell(s) does iron act as the anode?
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