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

14.4-14.9: Qualitatively Interpreting the Equilibrium Constant-Effect of a Catalyst

Qualitatively Interpreting the Equilibrium Constant:

-If K_c is a large number (example: 4.1x10⁸) the products are favored at equilibrium, and there will be more product than reactant. To obtain K_c, the reactants are in the denominator, and when you divide by a smaller and smaller number (as the reactants dwindle and the products emerge), the value of K_c will get bigger and bigger.

-If K_c is a very small number (example: 4.6x10^-31) the reactants are favored at equilibrium, and there will be more reactant than product. Again, to obtain K_c, the products are in the numerator, and when you divide a small number (products) by a larger number (reactants), the value of K_c will be very small.

-If K_c is within an order of magnitude near 1 (example: 7.2), there is an even distribution of reactants and products.

-If K_c = 0, there has been no reaction.

Predicting the Direction of Reaction:

Q_c = reaction quotient
-Obtained the same exact way used to obtain K_c. Plug in the concentrations of the products raised to the power of their coefficients, divided by the concentrations of the reactants raised to the power of their coefficients.



-The only difference is K_c is at equilibrium and Q_c is not. So we can use Q_c to see how to get to K_c.

-If Q_c > K_c, numerator is too big, need more reactant, reaction will shift to the left
-If Q_c < K_c, denominator is too big, need more product, reaction will shift to the right -If Q_c = K_c, already at equilibrium

Calculating Equilibrium Concentrations:

1. Set up a table of concentrations (Initial, Change, and Equilibrium expressions in x)
2. Substitute expressions in x for equilibrium concentrations into the equilibrium-constant equation.
3. Solve the equilibrium-constant equation for the values of the equilibrium concentrations.

Note: If you cannot get rid of a squared x term easily by simply taking the square root of everything (which will happen if your initial concentrations are the same), you will have to use the quadratic equation.

This will yield two answers, only one answer will be correct. If one is positive and the other is negative, the positive answer is the correct answer. If both are positive, plug the answers into your x expressions to see which one is realistically plausible (doesn't yield a negative number for any concentrations).

Changing Reaction Conditions:

-Le Chatelier's Principle: when a system in chemical equilibrium is disturbed by a change in temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change variable.

-Three ways to alter equilibrium concentration of a gaseous reaction mixture:
1. Changing concentrations
2. Changing the partial pressures of the products and reactants by changing the volume
3. Changing the temperature

-Concentration:
-When more reactant is added or product removed from the reaction vessel, the reaction will shift to the right to compensate for this gain/loss.
-When more product is added or reactant removed, the reaction will shift to the left to compensate. More reactants will be produced and amount of product will be subsequently decreased.

-Pressures/Volume:
-Pressure is directly proportional to moles of gas, so a decrease in the volume of a reaction vessel will cause the reaction to move in the direction that has a smaller number of moles of gas.
-An increase in volume will cause the reaction to move in the direction that has a larger number of moles of gas.
-When there are equal numbers of moles of gas on either side of the reaction (example: 2 total moles of reactant yields 2 total moles of product), then a change in pressure causes no shift in the reaction.

-Temperature:
-Must know the change in enthalpy of a reaction (ΔH) to know how temperature will shift the reaction.
-If ΔH is negative, the reaction is exothermic. Heat can be "imagined" to be a product of the reaction, and when you increase a product, the reaction will shift to the left to compensate. So adding heat to an exothermic reaction will cause it to shift to the left. And a temperature decrease will cause the reaction to shift to the right.
-If ΔH is positive, the reaction is endothermic. Heat can now be imagined to be a reactant. When you increase a reactant, the reaction will shift to the right to compensate. So adding heat to an endothermic reaction will cause it to shift to the right. And a temperature decrease will cause the reaction to shift to the left.

Effect of a Catalyst:

K_c is unaffected by a catalyst. The rate of reaction increases, but the final composition at equilibrium will be the same!

14.1-14.2: Chemical Equilibrium-The Equlibrium Constant

Chemical Equilibrium:

At dynamic equilibrium, Rate_f = Rate_r, the forward rate equals the reverse rate. Both sides are always reacting at a microscopic level at the same rate.

Chemical Equilibrium: the state reached by a reaction mixture when the rates of forward and reverse reactions have become equal.

In a Moles vs Time graph, notice the amounts of substances become constant at equilibrium:

In a Rate vs Time graph, notice that the forward rate is large at first but steadily decreases, while the reverse rate starts at zero and steadily increases, until both rates are equal at equilibrium:


Molar Composition:

Example:

For the reaction

CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

When 1.000 mol CO and 3.000 mol H₂ are placed in a 10.00 L vessel at constant temperature and allowed to come to equilibrium, the mixture is found to contain 0.387 mol H₂O. How many moles of each substance are present in the equilibrium mixture?

Think about the starting amounts in moles, which are known. The change in the amount of moles as the reaction proceeds is an unknown, so we can label it "x". Each reactant will decrease by a factor of x (the factor is indicated by the coefficient of that particular reactant) and each product will increase by a factor of x (also found by the coefficient). The resulting amounts can be found by combining our known information into equations and solving for x.

It is easiest to do this by making a table. I, C, and E indicate Initial, Change, and Equilibrium.


If the initial amounts of products isn't stated, it is assumed to be 0. Also, if we know each product (with a coefficient of 1) increases by x, and we know the resulting amount of that product, then we know x. So here, x = 0.387 mol.

If the problem were to ask for concentrations, divide the results by the volume given, in this case 10.00L.

The Equilibrium Constant:

General Form of an Equation:
aA + bB ⇌ cC + dD

Where capital letters are reactants and products and lowercase letters are coefficients of the balanced chemical equation.

-The equilibrium constant can be obtained by multiplying the concentrations of products, dividing by the product of the concentrations of reactants, and raising each concentration term to a power equal to the coefficient in the chemical equation. This is much easier to see when it's written out:



-K_c can also be denoted as K_eq
-K_c has no units
-Concentrations must be entered in mols/Liter (M), otherwise convert to these units
-Must have an equilibrium to find K_c
-Ignore concentrations of pure liquids and solids, they are constant. Another way of writing concentrations is density (same dimensions of mass over volume) which will remain the same no matter how much of a solid or liquid you have.
-Just use the concentrations of gaseous and aqueous reactants and products.

Example:

CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

K_c = [CH₄][H₂O] / [CO][H₂]³

Example:

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

K_c = [NH₃]² / [NH₂][H₂]³

Example:

2A (g) + B(s) + 7C (aq) ⇌ 14D (l) + 3/2E (g)

K_c = [E]^3/2 / [A]²[C]⁷

-Manipulating the balanced chemical equation will change K_c:

-If you multiply the coefficients of a reaction through by a number, A, then K_c = K_c^A
-Suppose you multiply a reaction through by 2, then you must square K_c
-If you flip a reaction around so the reactants are now the products and the products are now the reactants, then you get the reciprocal, K_c = 1 / K_c

The Pressure Equilibrium Constant, K_p:

-Won't use this until later when we study Thermodynamics, but it is relevant now.
-K_p is another version of K_c
-The values plugged into K_p are partial pressures converted to atm instead of concentrations.
-Related to K_c via the ideal gas law (PV = nRT → n/V = P/RT)
-In other words, the molar concentration of a gas equals its partial pressure divided by RT, which is constant at a given temperature.


Example:

For the reaction
CH₄ (g) + 2O₂ (g) ⇌ CO₂ (g) + 2H₂O (g)

K_c = [CO₂][H₂O]² / [CH₄][O₂]²

K_p = P_CO2 · P_H2O² / P_CH4 · P_O2²

-Sometimes K_c = K_p, but not always, it depends on the coefficients of the chemical equation.

K_c = K_p when the sum of reactant coefficients equals the sum of the product coefficients.

Example:

For the same reaction
CH₄ (g) + 2O₂ (g) ⇌ CO₂ (g) + 2H₂O (g)

1 + 2 = 1 + 2

K_c = K_p

Equation to relate K_p to K_c:

K_p = K_c (RT)^Δn

R = the ideal gas constant (0.0821 atm.L/mol.K)
T = Temperature in degrees Kelvin
Δn = sum of the coefficients of the products – the sum of the coefficients of the reactants.

Note: when the sum of the reactant coefficients equals the sum of product coefficients, Δn will be zero. Anything raised to the zero power is 1, so RT^Δn will equal 1, and K_p will equal K_c times 1, or just K_c.

Equilibrium Constant for the Sum of Reactions:

-If a given chemical equation can be obtained by taking the sum of other equations, the equilibrium constant for the given equation equals the product of the equilibrium constants of the other equations.
-Note: use Hess's Law

Example:

Equation 1:
CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)
Equation 2:
CH₄ (g) + 2H₂S (g) ⇌ CS₂ (g) + 4H₂ (g)
Hess's Law:
CO + 3H₂ + CH₄ + 2H₂S ⇌ CH₄ + H₂O + CS₂ + 4H₂
Equation 3:
CO (g) + 2H₂S (g) ⇌ H₂O (g) + CS₂ (g) + H₂ (g)

K_c for Equation 1 = K₁
K₁ = [CH₄][H₂O] / [CO][H₂]³

K_c for Equation 2 = K₂
K₂ = [CS₂][H₂]⁴ / [CH₄][H₂S]²

K_c for Equation 3 = K₁ · K₂
K₃ = [CS₂][H₂O][H₂] / [CO][H₂S]²
(K₃ can be verified by looking at Equation 3 and writing K_c directly from it)