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

19.2-19.4: Construction of Voltaic Cells-Cell Potential

Construction of Voltaic Cells:

Terms:

Electrochemical Cell: a system consisting of electrodes that dip into an electrolyte and in which a chemical reaction either uses or generates an electric current.

Voltaic/Galvanic Cell: an electrochemical cell in which a spontaneous reaction generates an electric current.

Electrolytic Cell: an electrochemical cell in which an electric current drives an otherwise non-spontaneous reaction.



Galvanic/Voltaic Cell:

A voltaic cell consists of two half-cells.


Half-cell: the portion of an electrochemical cell in which a half reaction takes place, electrically connected to the other half-cell.

Here, one is a zinc/zinc ion half-cell where a strip of zinc metal is in a solution of a zinc salt. The other is a copper metal strip in a solution of a copper salt. These are called a zinc electrode (the anode) and copper electrode (the cathode), respectively.

An external circuit must connect the two zinc and copper half-cells two allow electron flow between the metal electrodes, represented in the diagram by the placement of the voltmeter.

The zinc electrode is dissolved as it's oxidized (as the zinc metal atom loses two e^–). These e^– flow through the zinc electrode to the external circuit. From the external circuit, they reach the copper electrode and the other half-cell. This gives the copper electrode a negative charge, which attracts the positively charged copper cations in solution. Copper is then deposited on the copper electrode as copper cations become reduced to copper metal.

One metal must be able to reduce the cation of the other without allowing the two solutions to mix. The charges in the solution need to remain in contact without allowing the metals to react. The solution to this problem is a salt bridge, the peach-colored tube filled with KCl in the diagram.

Salt Bridge: a tube of an electrolyte in a gel that is connected to the two half-cells of a voltaic cell; the salt bridge allows the flow of ions but prevents the mixing of the different solutions that would allow direct reaction of the cell reactants.

The two half-cell reactions are:

Zn (s) → Zn²⁺ (aq) + 2e^–
(oxidation half-reaction, a species loses electrons)

Cu²⁺ (aq) + 2e^– → Cu (s)
(reduction half-reaction, a species gains electrons)

Anode: the electrode at which oxidation occurs. Electrons are given up by the anode and flow to the cathode. Represented with a negative sign to show that electrons flow from it.
Cathode: the electrode at which reduction occurs. Electrons flow from the anode to the cathode. Represented with a positive sign.
Cell Reaction: the sum of the two half-reactions, the net reaction that occurs in the voltaic cell.

Cell Reaction:
Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)

Note: electrons are never aqueous, don't give them phase labels.

To sketch and label a voltaic cell, it helps to remember the key information:
1. Remember the basic setup of two half-cells and their electrodes connected by an external current and an internal salt bridge.
2. Recognize any information in the problem that will help you figure out which half-cell reaction is the reduction and which is the oxidation. If one half-cell is known to be reduced, then we know that it is the cathode and the other is the anode.
3. Remember the electrons flow away from the anode to the cathode.


Notation:

Example:

For the voltaic cell in the above diagram, it is written as:
Zn (s) │ Zn²⁺ (aq) ║ Cu²⁺ (aq) │ Cu (s)

-The anode (oxidation half-cell) is written on the left, the cathode (reduction half-cell) is written on the right.
-The cell terminals are written on the far right and far left of the notation. The salt solutions are written on the inner right and inner left of the notation.
-The single vertical line represents a phase boundary, for example between the solid and the solution.
-The double vertical line represents the salt bridge.
-When the half reaction involves a gas, an inert material such as platinum may be used as a surface for the reaction. In this case, Pt will be added to the outside end of that electrode half-reaction.
-Coefficients are not included.

Example:

Write the following reaction in cell notation:

2Tl (s) + Sn²⁺ (aq) → 2Tl⁺ (aq) + Sn (s)

First write the half-reactions and find out what is reduced and what is oxidized.

2Tl (s) → 2Tl⁺ (aq) + 2e^– (oxidation)
2e^– + Sn²⁺ (aq) → Sn (s) (reduction)

Apply notation:

Tl (s) │ Tl⁺ (aq) ║ Sn²⁺ (aq) │ Sn (s)

Example:

Write the following reaction in cell notation:

Zn (s) + 2Fe³⁺ (aq) → Zn²⁺ (aq) + 2 Fe²⁺ (aq)
(with platinum used as a catalyst).

First write the half-reactions and find out what is reduced and what is oxidized.

Zn (s) → Zn²⁺ (aq) + 2e^–
Fe³⁺ (aq) + e^– → Fe²⁺ (aq)

Apply notation:

Zn (s) │ Zn²⁺ (aq) ║ Fe³⁺ (aq) , Fe²⁺ (aq) │ Pt

Note: There is a comma between Fe³⁺ and Fe²⁺ on the right side because they are both in the same phase, and platinum appears on the end.

To write a cell reaction from the cell notation, write each half-reaction from the notation and add them together.

Example:

Tl (s) │ Tl⁺ (aq) ║ Sn²⁺ (aq) │ Sn (s)

Write half-reactions:

2Tl (s) → 2Tl⁺ (aq) + 2e^–
2e^– + Sn²⁺ (aq) → Sn (s)

Take the sum of these reactions:

2Tl (s) + 2e^– + Sn²⁺ (aq) → 2Tl⁺ (aq) + 2e^– + Sn (s)
(Simplify)
2Tl (s) + Sn²⁺ (aq) → 2Tl⁺ (aq) + Sn (s)

Note: These cell notations show products and reactants in standard state:
aq → 1.00 M
gas → 1 atm
When not in standard state, the information for concentration or pressure goes in parentheses just to the right for each species.

Example:

Zn (s) │ Zn²⁺ (aq) ║ H⁺ (aq) (0.200 M) │ H₂ (g) │ Pt

Cell Potential:

Potential Difference: the difference in electric potential (electrical pressure) between two points. Measured in volts (V).

The electrical work expended on a charge moving through an electric potential:
Electrical work = charge x potential difference
Using corresponding units:
Joules = coulombs x volts → J = C x V

Faraday Constant: Denoted F, the magnitude of the charge on one mole of electrons; equal to 96,485 C/mol e^–. The faraday is a unit of charge equal to 96,485 C.

Cell Potential: the maximum potential difference between the electrodes of a voltaic cell. Denoted E_cell. E_cell must be positive for a spontaneous process. Larger values for E_cell are desirable. The larger the E_cell, the greater the energy we get out of the system.

Maximum work obtainable form a voltaic cell:

W_max = ∆G^o = –nFE_cell
where:
n = number of moles of electrons transferred in the overall cell equation.
F = the Faraday constant
E_cell = the cell potential
Note: ∆G^o will be less than zero for a spontaneous process.