Question
Standard Reduction (Electrode) Potentials at 25 oC Half-Cell Reaction Eo (volts) Standard Reduction (Electrode) Potentials at...
|
Standard Reduction (Electrode) Potentials at 25 oC |
|
|---|---|
| Half-Cell Reaction | Eo (volts) |
|
Standard Reduction (Electrode) Potentials at 25 oC |
|
| Half-Cell Reaction | Eo (volts) |
| F2(g) + 2 e- 2 F-(aq) | 2.87 |
| Ce4+(aq) + e- Ce3+(aq) | 1.61 |
| MnO4-(aq) + 8 H+(aq) + 5 e- Mn2+(aq) + 4 H2O(l) | 1.51 |
| Cl2(g) + 2 e- 2 Cl-(aq) | 1.36 |
| Cr2O72-(aq) + 14 H+(aq) + 6 e- 2 Cr3+(aq) + 7 H2O(l) | 1.33 |
| O2(g) + 4 H+(aq) + 4 e- 2 H2O(l) | 1.229 |
| Br2(l) + 2 e- 2 Br-(aq) | 1.08 |
| NO3-(aq) + 4 H+(aq) + 3 e- NO(g) + 2 H2O(l) | 0.96 |
| 2 Hg2+(aq) + 2 e- Hg22+(aq) | 0.920 |
| Hg2+(aq) + 2 e- Hg(l) | 0.855 |
| Ag+(aq) + e- Ag(s) | 0.799 |
| Hg22+(aq) + 2 e- 2 Hg(l) | 0.789 |
| Fe3+(aq) + e- Fe2+(aq) | 0.771 |
| I2(s) + 2 e- 2 I-(aq) | 0.535 |
| Fe(CN)63-(aq) + e- Fe(CN)64-(aq) | 0.48 |
| Cu2+(aq) + 2 e- Cu(s) | 0.337 |
| Cu2+(aq) + e- Cu+(aq) | 0.153 |
| S(s) + 2 H+(aq) + 2 e- H2S(aq) | 0.14 |
| 2 H+(aq) + 2 e- H2(g) | 0.0000 |
| Pb2+(aq) + 2 e- Pb(s) | -0.126 |
| Sn2+(aq) + 2 e- Sn(s) | -0.14 |
| Ni2+(aq) + 2 e- Ni(s) | -0.25 |
| Co2+(aq) + 2 e- Co(s) | -0.28 |
| Cd2+(aq) + 2 e- Cd(s) | -0.403 |
| Cr3+(aq) + e- Cr2+(aq) | -0.41 |
| Fe2+(aq) + 2 e- Fe(s) | -0.44 |
| Cr3+(aq) + 3 e- Cr(s) | -0.74 |
| Zn2+(aq) + 2 e- Zn(s) | -0.763 |
| 2 H2O(l) + 2 e- H2(g) + 2 OH-(aq) | -0.83 |
| Mn2+(aq) + 2 e- Mn(s) | -1.18 |
| Al3+(aq) + 3 e- Al(s) | -1.66 |
| Mg2+(aq) + 2 e- Mg(s) | -2.37 |
| Na+(aq) + e- Na(s) | -2.714 |
| K+(aq) + e- K(s) | -2.925 |
| Li+(aq) + e- Li(s) | -3.045 |
Use standard reduction potentials to calculate the equilibrium constant for the reaction:
2Fe3+(aq) + Ni(s)2Fe2+(aq) + Ni2+(aq)
Hint: Carry at least 5 significant figures during intermediate
calculations to avoid round off error when taking the
antilogarithm.
Equilibrium constant: __________
G° for this reaction would be _________(greater or less) than
zero.
Homework Answers
Answer #1
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