2 Fe 2+ (aq) + I 2(s) c. 2 Fe 2+ (aq) +Cl 2(g) ® 2 Fe 3+ (aq) + 2Cl - (aq) Solut

Question

2 Fe2+(aq) + I2(s)c. 2 Fe2+(aq) +Cl2(g) ® 2 Fe3+(aq) + 2Cl-(aq)

Explanation / Answer

NOTE, I will be using values of cell potential from my owntextbook. So your answers may vary slightly depending on what yourcell potential values are. In these problems you will need to look up the half-reactions andfind the cathode and anode. The cathode will be the half-reactionwith the higher cell potential. a) The half-reactions and their cell potentials are: Cathode: Cu2+(aq) + 2e- =>Cu(s)    0.34 V Anode: Fe2+(aq) + 2e- => Fe(s) -0.45V Eocell = Eocathode -Eoanode     = 0.34 - (-0.45) = 0.79 V b) The half-reactions and their cell potentials are: Cathode: Fe3+(aq) + e- =>Fe2+(aq)      0.77 V Anode: I2(s) + 2e- => 2I-(aq)    0.54 V Eocell = Eocathode -Eoanode     = 0.77 - (0.54) = 0.23 V c) The half-reactions and their cell potentials are: Cathode: Cl2(g) + 2e- => 2Cl-(aq)       1.36 V Anode: Fe3+(aq) + e- =>Fe2+(aq)      0.77 V Eocell = Eocathode -Eoanode     = 1.36 - (0.77) = 0.59 V Hope that helps! Let me know if I made a mistake.

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