# 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-

91% (311 ratings) ###### 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?