Electrochemical Reactions and Buffers

pH and Hydrolysis

The hydrolysis of $A^-$ produces $OH^-$ , increasing the pH to 8.8 at the equivalence point.
The acid is completely neutralized at the equivalence point, so it is not a buffer at that point.

Buffering Region

A buffer exists from the start of neutralization until the neutralization is complete.
Adding $OH^-$ doesn't fully neutralize the acid, leaving both weak acid and conjugate base.
At half the equivalence point:
- Moles of acid neutralized equals the moles of $OH^-$ . added.
- Half of the acid remains, with an equal amount of conjugate base ( $A^-$ ).
- $pH = pK_a$

Henderson-Hasselbalch Equation and pH

Before half the equivalence point:
- Added base neutralizes acid, but acid concentration is greater.
- pH < pK_a
- $pH = pK_a + \log\frac{[A^-]}{[HA]}$
- Moles of $HA$ > moles of $A^-$ .
After half the equivalence point:
- More acid is neutralized, creating more $A^-$ .
- pH > pK_a
- At half equivalence point, moles of acid = moles of conjugate base, so $pH = pK_a$

Determining Volume of NaOH

Problem: Determine the volume of 0.1 M NaOH needed to add to 100 mL of 0.1 M HNO2 to reach a pH of 3.4.
pH of 3.4 corresponds to half the equivalence point.
Initial moles of $HNO_2$ : $100 \text{ mL} \times 0.1 \text{ M} = 0.01 \text{ moles}$
Moles of $OH^-$ needed: $\frac{0.01}{2} = 0.005 \text{ moles}$
Reaction:
$HNO2 + OH^- \rightarrow NO2^- + H_2O$
At half equivalence:
$[HNO2] = [NO2^-] = 0.005 \text{ moles}$

$pH = pK_a = 3.4$

Volume of 0.1 M NaOH needed: $\frac{0.005 \text{ moles}}{0.1 \text{ M}} = 0.05 \text{ L} = 50 \text{ mL}$

Strong Base and Weak Acid

Adding a strong base like NaOH to a weak acid does not always create a buffer.
Equimolar proportions lead to complete neutralization, eliminating the buffer.
Adding half the moles creates a buffer at half the equivalence point, where
$pH = pK_a$

Preparing a Buffer with a Specific pH

To prepare a buffer with a pH of 3, add less NaOH to avoid complete neutralization.
More acid (HNO2) than its conjugate base (NO2-) is required when pH < pKa.
Use math to justify the quantities.

$pH = pKa + \log\frac{[NO2^-]}{[HNO_2]}$

Example

If the desired pH = 4 (where pKa = 3.4), more $NO_2^-$ is needed.
Given: 0.01 moles of $HNO_2$
To achieve pH 4, the concentration of $NO2^-$ must be greater than $HNO2$

Buffering Capacity

Comparing two buffers: 0.1 M $HNO2$ and 1 M $HNO2$ .
The buffer with 1 M $HNO_2$ has a higher buffering capacity towards a strong base.
More concentrated buffer resists pH changes by neutralizing more strong base.

Buffers: Definition and Function

Buffers consist of a weak acid and its conjugate base (acidic buffer) or a weak base and its conjugate acid (basic buffer).
They maintain solution pH by neutralizing small amounts of strong acid or base.
A buffer can also be created by adding strong acid to produce a base up to the neutralization point.

Buffering Capacity

Buffering capacity is higher with increased concentrations of the acid or its conjugate.
Titration graphs show a gradual pH change initially due to the buffer's neutralizing effect.
The region where the curve is parallel to the x-axis represents the buffering region.

Determining pH

Determine whether pH will be greater than, less than, or equal to pKa.
Equation:
$pH = pK_a + \log\frac{[base]}{[acid]}$
Example 1:
If [acid] > [base], then pH < pKa If [acid] = 10 and [base] = 6, then $pH = pKa + \log\frac{6}{10}$
$pH = pKa + \log(0.6)$ $pH < pKa$

Electrochemistry: Spontaneous vs. Non-Spontaneous Cells

Check for spontaneity:
- Spontaneous cells produce electricity without an external power source.
- Non-spontaneous cells require a power source.
  Identifying Non-Spontaneous Cells
- If there is a power source, the $E^0$ will be negative.
- Anions travel to the anode and cations migrate to the cathode. Example of Non-Spontaneous Cell
  - Lithium production made by electrolyzing molten lithium chloride.
    Overall equation checks
  - When given equations, verify more negative will be the anode.
    Spontaneous Cells: Characteristics
  - $E^0$ is positive, and $\Delta G^0$ is negative.

Reactivity and Reduction Potential

More negative $E^0$ : more reactive metal
Lithium: produces maximum electricity in spontaneous cells; must be the anode

Electrolysis and Water

Water interference: Electrolyzing molten lithium chloride must be devoid of any water!

Water's Role
*Water will interfere with lithium metal production.

Electron Flow

Electrons
*Electron flow occurs always from the anode --> cathode.

Non-Spontaneous Cells

Anions
*Anions yield electrons and transform into the halogen, such as $Cl_2$ .

Oxidation and Reduction Review

Oxygen Oxidation States
* Oxygen general oxidation state: -2, except when binding to Fluorine.
* Oxygen is only positive when reacted with fluorine.
Hydrogen Oxidation States
* Hydrogen is +1 with nonmetals and -1 with metals.

Cell Potential Calculation

$E_{cell}^0 = anode + cathode$
Make sure to calculate decimals properly.

Calculating Delta G

$\Delta G = -nFE$

Stoichiometric Calculation

Question: Calculating moles of substance from electrolytic process.
Steps

I = q/t, where q is in coulombs.
Bring in Faraday's constant in order to relate it to the current.
96,485 coulombs = 1 mole electrons.

Understanding Electrochemical Half-Reactions

At Electrode
*Always base reduction/oxidation process on the particular reaction and # of electrons
*Magnesium required 2 electrons, while MNO2 only takes 1.

Voltic Vs Electrolytic Cells

Understand that the number of electrons always equals what the stoichiometry is.

Calculate Delta G from Here

Bring equations equal for delta g.

Understanding Batteries: AA vs. D cells

Battery Chemistry is Identical
*Double AA = D cell (chemistry), both possess same cell potential. But, D cell may run longer.
*To increase potential, you must increase reactants.

Nernst Equation

$E = E^0 - \frac{RT}{nF} \ln Q$

The Salt Bridge (Voltaic Cells)

Imbalance of Charges Disrupts Flow
*Salt helps assist with balancing charge and prolong electrical production.

Charge Conservation
*Must conserve charges by the electrodes for electrical productivity. Anodes --> chamber, therefore negative charge neutralizes, so positive charges are there to balance.

Thermodynamics

Role within Electronegativity
*Enthalpy, and entropy's product helps dictate spontaneity of reaction (Gibbs Free Energy)
* $\Delta H$ (negative) favored; exothermic.
* $\Delta S$ (positive) favored.
*Solid --> liquid --> gas (more dispersion is entropy + ).
Equations
* $\Delta G=\Delta H-T\Delta S$

When temperature = equilibrium, then K = 1.
When $\Delta G$ = negative then: K>1, then: when $\Delta G$ =zero then K = 1.