The information from the Table of Standard Reduction Potentials can be used to predict chemical reactivity. If two half-reactions can be combined to give a positive cell voltage, the reaction will spontaneous.
Will Br2(l) spontaneously oxidize Fe2+(aq)? If so, what is the cell reaction and what is the standard cell potential?
From the Table of Standard Reduction Potentials we find the following:
Fe3+(aq) + e– → Fe2+(aq) E° = +0.77 V
Br2(l) + 2 e– → 2 Br–(aq) E° = +1.07 V
The question asks if Fe2+(aq) can be oxidized, so the reaction we are interested in is:
Fe2+(aq) → Fe3+(aq) + e–
with E°oxidation = –0.77 V
We now have both an oxidation and a reduction so a net reaction may occur as:
2 Fe2+(aq) + Br2(l) → 2 Fe3+(aq) + 2 Br–(aq)
The cell potential for this reaction is
E° = E°reduction + E°oxidation = +1.07 + (–0.77) = +0.30 V
Since E° > 0, the reaction is spontaneous so Br2(l) will oxidize Fe2+(aq).
Note: we do not use stoichiometry when adding half-reaction potentials to find net cell potentials. (This may be the only time you don't have to use stoichiometry in chemical problems; even when using half-cell potentials to find other half-cell potentials, stoichiometry must be used!)
Will I–(aq) reduce Cr3+(aq) to the free metal. If so, what is the net reaction and what is the standard cell potential?
From the Table of Standard Reduction Potentials, we find:
Cr3+(aq) + 3 e– → Cr(s) E° = –0.74 V
I2(s) + 2 e– → 2 I–(aq) E° = +0.54 V
The proposed reaction is:
2 Cr3+(aq) + 6 I–(aq) → 2 Cr(s) + 3 I2(s)
The cell potential for this reaction is
E° = (–0.74) + (–0.54) = –1.28 V
E° < 0 so the reaction is nonspontaneous, i.e.
Cr3+(aq) + I–(aq) → NR
The Gibb's Energy allows us to connect cell potentials to equilibrium constants:
ΔG° = –nFE°
ΔG° = –RTln Keq
–nFE° = –RTln Keq
E° = RT nF ln Keq
R is the gas constant = 8.314 J/mol·K
T is the temperature in units of K
F is Faraday's constant = 96485 coul/mol
n is the number of electrons transferred in the balanced equation
E° is the cell potential at standard conditions
Keq is the thermodynamic equilibrium constant
This relationship allows us to find equilibrium constants from cell potentials or cell potentials from equilibrium constants!
Find Keq at 25 °C for the reaction
Sn4+(aq) + Cu+(aq) → ← Sn2+(aq) + Cu2+(aq)
Strategy: balance the reaction; determine the cell potential from standard potentials; use the relationship between cell potentials and equilibrium constants.
Sn4+(aq) + 2 e– → Sn2+(aq) E°red = +0.154 V
Cu+(aq) → Cu2+(aq) + e– E°ox = –E°red = –(0.159) V
Sn4+(aq) + 2 Cu+(aq) → ← Sn2+(aq) + 2 Cu2+(aq) E° = E°red + E°ox = 0.154 + (–0.159) = –0.005 V
Use the standard potential
With T = 25 + 273 = 298 K
R = 8.314 J/mol·K
F = 96485 coul/mol
n = 2
–0.005 = (8.314)(298) (2)(96485) ln Keq
ln Keq = –0.4
Keq = e–0.4 = 0.7
Find Ksp for AgCl at 25 °C using electrochemical data.
The reaction of interest is:
AgCl(s) → ← Ag+(aq) + Cl–(aq)
We must find an oxidation and a reduction that add up to the desired reaction. Inspection of the Table of Reduction Potentials shows two reactions that involve silver:
Ag+(aq) + e– → Ag(s) E°red = +0.800 V
AgCl(s) + e– → Ag(s) + Cl–(aq) E°red = +0.2223 V
If we write the silver ion reduction as an oxidation:
Ag(s) → Ag+(aq) + e– E°ox = –0.800 V
AgCl(s) + e– → Ag(s) + Cl–(aq) E°red = +0.2223 V
Then, the sum of the oxidation and reduction is:
AgCl(s) → ← Ag+(aq) + Cl–(aq) E° = +0.2223 + (–0.800) = –0.578 V
Now we can use E° = –0.578
With T = 25 + 273 = 298 K
R = 8.314 J/mol·K
F = 96485 coul/mol
n = 1
–0.578 = (8.314)(298) (1)(96485) ln Keq
Keq = Ksp, so
ln Ksp = –22.5
Ksp = e–22.5 = 1.7×10–10
(This compares to 1.8×10–10 from the Table of Solubility Product Constants.)