Problem: 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:Cell 1: a 1.0 M CuCl2(aq) solution with a Cu electrodeCell 2: a 1.0 M NiCl2(aq) solution with a Ni electrodeCell 3: a 1.0 M ZnCl2(aq) solution with a Zn electrodeIn which voltaic cell(s) does iron act as the anode?

FREE Expert Solution

• ↓ 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-

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Problem Details

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:


  • Cell 1: a 1.0 M CuCl2(aq) solution with a Cu electrode
  • Cell 2: a 1.0 M NiCl2(aq) solution with a Ni electrode
  • Cell 3: a 1.0 M ZnCl2(aq) solution with a Zn electrode


In which voltaic cell(s) does iron act as the anode?


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