Two half-cells are shown below: Fe2+(aq) + 2e- ↔ Fe(s) …
Two half-cells are shown below: Fe2+(aq) + 2e- ↔ Fe(s) Ecell = – 0.44 V Cu2+(aq) + 2e- ↔ Cu(s) Ecell = +0.34 V Hint: The half-cell with the highest positive potential gets reduced. the other half-cell must get oxidized. The complete cell reaction is
Read DetailsConsider the following half-cell reactions: Zn2+(aq) + 2e-…
Consider the following half-cell reactions: Zn2+(aq) + 2e- ↔ Zn(s) Ecell = -0.76 V Cu2+(aq) + 2e- ↔ Cu(s) Ecell = +0.34 V Identify the species (element or ion) that is acting as the oxidizing agent Hint: The half-cell with the highest positive potential gets reduced
Read DetailsTwo half-cell reactions are shown below: Fe2+(aq) + 2e- …
Two half-cell reactions are shown below: Fe2+(aq) + 2e- ↔ Fe(s) Ecell = – 0.44 V Cu2+(aq) + 2e- ↔ Cu(s) Ecell = +0.34 V Hint: The half-cell with the highest positive potential gets reduced. the other half-cell must get oxidized. Find the Gibbs free energy for the above cell reaction (ΔG = -nFE: F = 96485 C, n = no. of electrons and E = cell potential). Use the cell voltage you received in questions 21.
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