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 standard free energy change in kJ for the reaction:
Cd2+(aq) + 2Cr2+(aq)Cd(s) + 2Cr3+(aq)
Answer:________ kJ
K for this reaction would be _________(greater or less) than
one.
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