19.4-19.7: Cell Potential-Dependence of Cell Potential on Concentration

W_max = ∆G^o = –nFE_cell

Example:

What is the W_max that can be obtained from 351g of zinc metal in the galvanic cell with the overall cell reaction of

Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)

W_max = ∆G^o = –nFE_cell

n = moles of e^– transferred = 2
F = 96485 C / mol e^–
to find the W_max we need E_cell.

To find E_cell, reference Table 19.1: Standard Electrode (Reduction) Potentials(E^o), page 786.

First, write the two half-reactions from the overall cell reaction:
Zn (s) → Zn²⁺ (aq) + 2e^– (oxidation half-reaction)
Cu²⁺ (aq) + 2e^– → Cu (s) (reduction half-reaction)

This table gives E^o values only for the reduction half-reactions. In order to get the E^o value for an oxidation half-reaction, reverse the order to find it as it is written in the table, and take the negative of the given E^o number given for that reaction.

Zn (s) → Zn²⁺ (aq) + 2e^– is not a reaction that is in Table 19.1 because it is an oxidation reaction. Reverse the reaction to find it as it is written in the table:
Zn²⁺ (aq) + 2e^– → Zn (s)

The given E^o value is – 0.76 V. Since the reaction we are working with is the reverse of this one, take the negative of this value. So:

Zn (s) → Zn²⁺ (aq) + 2e^– (E^o = + 0.76 V)
Cu²⁺ (aq) + 2e^– → Cu (s) (E^o = + 0.34 V)

To get the value for E_cell for the overall reaction, add these two E^o values together.

0.76 V + 0.34 V = 1.10 V = E_cell

Now back to the original question:

W_max = ∆G^o = –nFE_cell
W_max = –(2)(96485)(1.10) = –212 kJ per mole of Zn

Remember that the problem stated that we had 351g of Zn, not one mole of Zn, so:

(351g Zn) x (1 mol Zn / 65.4g Zn) x (–212 kJ / 1 mol Zn) = 1.14 x 10³kJ



Standard Cell Potential: Denoted E^o_cell, it is the cell potential of a voltaic cell operating under standard-state conditions (solute concentrations are each 1 M, gas pressures are each 1 atm, and the temperature has a specified value, usually 25^oC.) The degree sign signifies standard-state conditions.

Standard Electrode Potential: Denoted E^o, it is the electrode potential under standard-state conditions. The degree sign signifies standard-state conditions.

Standard electrode potentials help determine strengths of oxidizing and reducing agents.
-The strength of reducing agents increases going up the table.
-Example: Lithium is the strongest reducing agent.
-The strength of oxidizing agents increases going down the table.
-Example: F₂ is the strongest oxidizing agent.

Example:

Which is the stronger reducing agent? Li (s) or Li⁺?

Li, it has more e^– to give.

To determine the direction of spontaneity with this information as well:
-The stronger oxidizing agent will be on the reactant side when the equation is written as a spontaneous reaction.
-The stronger reducing agent will be on the reactant side of the spontaneous reaction.


Equilibrium Constants from Cell Potentials:

If W_max = ∆G^o
and
∆G^o = –nFE^o_cell
and
∆G^o = –RtlnK
then
–nFE^o_cell = –RtlnK
nFE^o_cell = RtlnK
Since we're at standard conditions, we can rearrange the equation and plug in known values to get:
E^o_cell = (0.0592 / n)logK

Example:

Find K for the overall reaction:
2H³⁺ (aq) + 3Fe (s) → 2Al (s) + 3 Fe²⁺ (aq)

E^o_cell = (0.0592 / n)logK
To find K, need n and E^o_cell. So start, like always, with writing the half-reactions:

2Al³⁺ (aq) + 6e^– → 2Al (s) (E^o = –1.66 V)
3Fe (s) → 3Fe²⁺ (aq) + 6e^– (E^o = +0.41 V)
E^o_cell = –1.66 V + 0.41 V = –1.25 V
n = 6

(–1.25 V) = (0.0592 / 6)(logK)
logK = –127
K = 10^–127

Nernst Equation:

What if we're at non-standard conditions (assuming temperature remains at 298K)? Luckily, the monopoly guy derived an equation that relates cell potential to standard cell potential.


Walther Hermann Nernst

E_cell = E^o_cell – (0.0592 / n)(logQ)
where Q is the reaction quotient (remember this is the same as K except not at equilibrium).

Example:

Zn (s) │Zn²⁺ (0.155M) ║ Ag⁺ (0.0200M) │ Ag (s)
Find E_cell. (Must use Nernst equation because it is at non-standard conditions.)

E_cell = E^o_cell – (0.0592 / n)(logQ)

First convert notation into half-reactions and find E^o_cell:
Zn (s) → Zn²⁺ (aq) + 2e^– (E^o = +0.76 V)
2Ag⁺ (aq) + 2e^– → 2Ag (s) (E^o = +0.80 V)
E^o_cell = +1.56 V
n = 2

Add the half-reactions to get the overall reaction:
Zn (s) + 2Ag⁺ (aq) → Zn²⁺ (aq) + 2Ag (s)

Write Q using the same format used to write K:
Q = [Zn²⁺] / [Ag⁺]²
Q = (0.155) / (0.0200)²
Q = 387.5

E_cell = E^o_cell – (0.0592 / n)(logQ)
E_cell = (1.56) – (0.0592 / 2)(log 387.5)
E_cell = 1.48 V

Determination of pH:

A pH meter actually measures E_cell and then backtracks to pH.

pH = (0.76 – E_cell) / (0.0592)

Example:

If the pH = 4.3, what is the E_cell of the test system?
4.3 = (0.76 – E_cell) / (0.0592)
E_cell = 0.505 V