Chemistry 112

Other uses of Standard Cell Potentials

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.

Examples

Will Br2(l) spontaneously oxidize Fe2+(aq)? If so, what are the cell reaction and the standard cell potential?

From the Table of Standard Reduction Potentials we find the following:

 Fe3+(aq) + e–Fe2+(aq) Eo = +0.77 V Br2(l) + 2 e–2 Br–(aq) Eo = +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 Eooxidation = –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

Eo = Eoreduction + Eooxidation = +1.07 + (–0.77) = +0.30 V

Since Eo > 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 are the net reaction and the standard cell potential?

From the Table of Standard Reduction Potentials, we find:

 Cr3+(aq) + 3 e–Cr(s) Eo = –0.74 V I2(s) + 2 e– 2 I–(aq) Eo = +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

Eo = (–0.74) + (–0.54) = –1.28 V

Eo < 0 so the reaction is nonspontaneous, i.e.

Cr3+(aq) + I(aq)NR

Using Electrochemical Measurements to Find Equilibrium Constants

The Gibb's Energy allows us to connect cell potentials to equilibrium constants:

Go = –nFEo

Go = –RTln Keq

–nFEo = –RTln 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

Eo 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!

Examples

Find Keq at 25 oC 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) Eored = 0.154 V Cu+(aq)Cu2+(aq) + e– Eoox = –Eored = –(0.159) V Net: Sn4+(aq) + 2 Cu+(aq)Sn2+(aq) + 2 Cu2+(aq) Eo = Eored + Eoox = 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

ln Keq = –0.4

Keq = e–0.4 = 0.7

Find Ksp for AgCl at 25 oC 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) Eo = +0.800 V AgCl(s) + e–Ag(s) + Cl–(aq) Eo = +0.2223 V

If we write the silver ion reduction as an oxidation:

 Ag(s)Ag+(aq) + e– Eoox = –0.800 V

Then, the sum of the oxidation and reduction is:

 AgCl(s) Ag+(aq) + Cl–(aq) Eo = +0.2223 + (–0.800) = –0.578 V

Now we can use

With T = 25 + 273 = 298 K

R = 8.314 J/mol·K

F = 96485 coul/mol

n = 1

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.)