CellPotential 1 (Points: 8.33) Using table 20.1, calculate the E°cell for the fo
ID: 689653 • Letter: C
Question
CellPotential 1
(Points: 8.33) Using table 20.1, calculate the E°cell for the followingreaction:2Fe2+(aq) + Cu2+(aq) --->2Fe3+(aq) + Cu(s)
1. ) None of them
2. ) -0.43 V
3. ) +0.43 V
4. ) -0.78V
5. ) +0.78 V
CellPotential 1
(Points: 8.33) Using table 20.1, calculate the E°cell for the followingreaction:2Fe2+(aq) + Cu2+(aq) --->2Fe3+(aq) + Cu(s)
1. ) None of them
2. ) -0.43 V
3. ) +0.43 V
4. ) -0.78V
5. ) +0.78 V
Explanation / Answer
Here our half reactions and reduction potentials are: Fe3+ + e- -->Fe2+ Eo =+0.77 Cu2+ + 2e- -->Cu Eo = +0.340 From Electrochemistry we know: Eocell = Ecathode -Eanode The cathode is the cell in which reduction potential is the largest(the iron cell in this problem), and anode is the smaller reduction potential (copper sln). So now we can equateEcell: Ecell = (0.77 - 0.34) Eocell = +0.430 V Hope this helped :)
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