Electrochemistry

Electrochemistry Overview

Voltaic vs. Electrolytic Cells

• Voltaic (Galvanic) Cells:
  • Produce energy spontaneously.
  • Cell potential ($E$) must be positive (E > 0).
  • Example: Battery powering a cell phone.
• Electrolytic Cells:
  • Require an external energy source to drive the reaction.
  • Cell potential can be positive or negative.
  • Example: Charging a battery, where energy flows from the outlet to the battery.

Voltaic Cell Example: Zinc and Copper

• Standard Reduction Potentials:
  • Zinc: $Zn^{2+}(aq) + 2e^- \rightarrow Zn(s)$, $E° = -0.76 V$
  • Copper: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$, $E° = +0.34 V$
• To obtain a positive cell potential (spontaneous reaction), reverse the zinc half-reaction:
  • $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$, $E° = +0.76 V$
• Net Reaction: Zinc + Copper(II) ions
  • $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$
  • Cell potential: $E° = 0.76 V + 0.34 V = 1.1 V$
• Standard Conditions:
  • Ion concentrations must be 1 molar (1 mol/L).
  • If concentrations differ, the cell potential will vary.

Oxidation and Reduction

• Oxidation:
  • Loss of electrons.
  • Increase in oxidation state.
  • Electrons on the right side of the half-reaction.
  • Example: Zinc losing two electrons ($Zn \rightarrow Zn^{2+} + 2e^-$).
• Reduction:
  • Gain of electrons.
  • Decrease in oxidation state.
  • Electrons on the left side of the half-reaction.
  • Example: Copper(II) ions gaining two electrons ($Cu^{2+} + 2e^- \rightarrow Cu$).

Electrochemical Cell Process

• Zinc atoms in the zinc electrode lose two electrons and enter the solution as $Zn^{2+}$ ions.
• Electrons flow from the zinc electrode (anode) towards the copper electrode (cathode).
• Copper(II) ions in the solution are attracted to the electrons at the copper electrode.
• $Cu^{2+}$ ions pick up two electrons and deposit themselves as copper atoms on the electrode.

Anode vs. Cathode

• Anode:
  • Zinc electrode where oxidation occurs.
  • Loses mass as zinc atoms leave the electrode and enter the solution.
  • Electrons flow from the anode
• Cathode:
  • Copper electrode where reduction occurs.
  • Gains mass as copper ions deposit themselves onto the electrode

Salt Bridge

• Purpose: To maintain charge balance.
• Ions travel through the salt bridge to maintain a stable electrical current.
• Cations:
  • Positively charged ions.
  • Travel toward the cathode.
• Anions:
  • Negatively charged ions.
  • Travel toward the anode.
• Example compounds for aqueous solutions:
  • Zinc electrode: Zinc sulfate ($ZnSO_4$) is most suitable due to its solubility.
  • Avoid insoluble compounds like zinc carbonate or zinc hydroxide.

Oxidizing and Reducing Agents

• Oxidizing Agent:

  • Causes oxidation by accepting electrons.
  • It is itself reduced.
  • Example: $Cu^{2+}$ ions.

• Reducing Agent:

  • Causes reduction by donating electrons.
  • It is itself oxidized.
  • Example: Zinc metal.

• The substance that is oxidized and reduced is always a reactant.

• Metals are usually good reducing agents.

• Metal ions can be oxidizing agents.

• Non-metals (e.g., fluorine, oxygen gas) are usually oxidizing agents.

Gibbs Free Energy and Cell Potential

• Relationship: $\Delta G = -nFE$
  • $\Delta G$: Gibbs free energy.
  • $n$: Number of moles of electrons transferred in the balanced reaction.
  • $F$: Faraday's constant (96485 coulombs per mole of electrons).
  • $E$: Cell potential (volts).
  • Under standard conditions: $\Delta G° = -nFE°$
• Example:
  • For the zinc-copper cell, $n = 2$, $E° = 1.1 V$
  • $\Delta G° = -2 * 96485 \frac{C}{mol} * 1.1 \frac{J}{C} = -212,267 \frac{J}{mol}$
  • This is the electrical work that the cell can do.

Spontaneity and Equilibrium Constant (K)

• Spontaneity:
  • Positive cell potential (E > 0) indicates a spontaneous reaction.
  • Negative cell potential (E < 0) indicates a non-spontaneous reaction.
  • Negative $\Delta G$ indicates a spontaneous reaction; positive $\Delta G$ indicates a non-spontaneous reaction.
• Equilibrium (K):
  • Positive $E$ and negative $\Delta G$: product-favored at equilibrium; K >> 1.
  • Negative $E$ and positive $\Delta G$: reactant-favored at equilibrium; K << 1.
• K is the ratio of products to reactants at equilibrium.

Calculating K from $\Delta G$

• Equation: $\Delta G = -RT \ln{K}$
  • $R$: Ideal gas constant (8.3145 J/mol*K when $\Delta G$ is in joules).
  • $T$: Temperature in Kelvin.
• Rearranging: $K = e^{\frac{-\Delta G}{RT}}$
• Example:
  • $\Delta G = -212,267 J/mol$, $T = 298 K$
  • $K = e^{\frac{-(-212,267)}{(8.3145 * 298)}} = 1.67 * 10^{37}$
  • Large K indicates a product-favored reaction.

Cell Notation

• Represents the electrochemical cell.
• Oxidation (anode) is on the left, and reduction (cathode) is on the right.
• Single line (|) separates different phases.
• Double line (||) represents the salt bridge.
• Example:
  • For the reaction $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$
  • Cell Notation: $Zn(s) | Zn^{2+}(aq) || Cu^{2+}(aq) | Cu(s)$

Nernst Equation (Non-Standard Conditions)

• Equation: $E = E° - (\frac{0.0591}{n}) \log{Q}$
  • $E$: Cell potential under non-standard conditions.
  • $E°$: Standard cell potential.
  • $n$: Number of moles of electrons transferred.
  • $Q$: Reaction quotient (ratio of products to reactants at initial conditions).
• Solids and liquids are not included in the expression for Q.
• Example:
  • If $\left[Zn^{2+}\right] = 0.1 M$ and $\left[Cu^{2+}\right] = 10 M$
  • $Q = \frac{\left[Zn^{2+}\right]}{\left[Cu^{2+}\right]} = \frac{0.1}{10} = 0.01$
  • $E = 1.1 - (\frac{0.0591}{2}) \log{0.01} = 1.13 V$

Le Chatelier's Principle and Cell Potential

• Increasing the concentration of reactants shifts the reaction to the right, making it more spontaneous, thus increasing the cell potential.
• Decreasing the concentration of products also shifts the reaction to the right, increasing the cell potential.
• Increasing products or decreasing reactants has the reverse effect.

Calculating Q from Non-Standard Cell Potential

• Rearranging the Nernst equation to solve for Q:
  • $E = E° - (\frac{0.0591}{n}) \log{Q}$
  • $\log{Q} = \frac{n (E - E°)}{-0.0591}$
  • $Q = 10^{\frac{n (E - E°)}{-0.0591}}$
• Example:
  • If $E = 0.97 V$, then $Q = 10^{\frac{2 (0.97 - 1.1)}{-0.0591}} = 25080$

Impact of Q on Cell Potential

• A large Q means a high ratio of products to reactants, which decreases the cell potential.
• A small Q means a high ratio of reactants to products, which increases the cell potential.

Standard vs. Non-Standard Values

• The non-standard cell potential (E) and non-standard Gibbs free energy ($\Delta G$) change with the concentration of reactants and products.
• The standard cell potential ($E°$), standard Gibbs free energy ($\Delta G°$), and equilibrium constant (K) remain constant if the temperature is constant.
• Equilibrium constant (K) depends only on temperature.

Stoichiometry Problems in Electrochemistry

• Key Equations:
  • $Q = I * t$
    • Q = charge in coulombs
    • I = current in amperes (amps)
    • t = time in seconds
  • 1 coulomb = 1 amp * 1 second
  • Faraday's constant: 96485 coulombs per mole of electrons.
  • Amp * seconds can be used in place of coulombs in calculations.

Example 1: Mass Deposited on Cathode

• Problem: A current of 1.25 amps passes through a solution of copper sulfate for 39 minutes. Calculate the mass of copper deposited on the cathode.
• Solution:
  1. Convert minutes to seconds: $39 \min * 60 \frac{s}{\min} = 2340 s$
  2. Use $Q = I*t$ to determine coulombs and then convert to moles of electrons $\frac{(2340 s * 1.25 A)}{96485 \frac{C}{mol e^-}}$
  3. Use the half-reaction $Cu^{2+} + 2e^- \rightarrow Cu$ with the mole ratio $\frac{1 mol Cu}{2 mol e^-}$
  4. Convert moles of copper to grams using the atomic mass of copper (63.55 g/mol).
  5. Calculation: $\frac{(39 \min * 60 \frac{s}{\min} * 1.25 A * 63.55 \frac{g}{\text{mol}})}{(96485 \frac{C}{mol e^-} * 2 e^-)} = 0.963 g$

Example 2: Calculating Average Current

• Problem: The mass of the zinc anode decreased by 1.43 grams in 56 minutes. Calculate the average current that passed through the solution.
• Solution:
  1. Convert grams of zinc to moles using the molar mass of zinc (65.39 g/mol).
  2. Use the half-reaction $Zn \rightarrow Zn^{2+} + 2e^-$ with the mole ratio $\frac{2 mol e^-}{ 1 mol Zn}$
  3. Convert moles of electrons to amps * seconds using Faraday's constant.
  4. Divide by the time in seconds (56 minutes * 60 seconds/minute) to find the current in amps.
  5. * Calculation: $\frac{(1.43 g / \frac{65.39 g}{mol} * 2 * \frac{96485 A*s}{mol})}{ (56 \min * \frac{60 s}{min}}= 1.226 A$

Example 3: Calculating Time for Deposition

• Problem: How long will it take in hours for a current of 745 milliamps (0.745 amps) to deposit 8.56 grams of chromium onto the cathode using a solution of chromium(III) chloride?
• Solution:
  1. Convert grams of chromium to moles using the molar mass of chromium (52 g/mol).
  2. Use the half-reaction $Cr^{3+} + 3e^- \rightarrow Cr$ with the appropriate mole ratio.
  3. Convert moles of electrons to amps * seconds using Faraday's constant.
  4. Divide by the current in amps (0.745 A) and convert seconds to hours.
  5. * Calculation: $\frac{8.56 g}{( \frac{52 g}{mol} * 3 * \frac{96485 A*s }{mol}) / (0.745 A * \frac{1 h}{3600 s})}= 17.77 hours$

Determining Strongest Oxidizing and Reducing Agents

• Metals are usually reducing agents (they lose electrons).
• Metal cations are oxidizing agents (they gain electrons).
• Non-metals are usually oxidizing agents.
• Non-metal anions are reducing agents.
• The strongest reducing agent is the metal that wants to give away electrons the most (most positive cell potential when the reaction is reversed).
• The strongest oxidizing agent is the metal cation that wants to acquire electrons the most (most positive cell potential).

Example 1: Metals and Metal Cations

• Given reduction potentials for Aluminum, Iron, Copper, and Silver.
• Aluminum is the strongest reducing agent because it has the most positive cell potential when the reaction is reversed ($Al \rightarrow Al^{3+} + 3e^-$).
• $Ag^+$ is the strongest oxidizing agent because it has the highest cell potential ($Ag^+ + e^- \rightarrow Ag$).

Example 2: Metals and Metal Cations

• Reducing Agents tend to give away electrons.
• Given reduction potentials for Magnesium, Zinc, and Lead.
• Magnesium (Mg) is the strongest reducing agent among the metals because its oxidation reaction from solid Mg to Mg2+ has the most positive standard potential
• Lead metal cation ($Pb^{2+}$) is the strongest oxidizing agent (least negative cell potential).

Non-Metals and Non-Metal Anions

• Neutral metals are reducing agents.
• Neutral non-metals are oxidizing agents.
• Metal cations are oxidizing agents.
• Non-metal anions are reducing agents.
• To Identify the strongest oxidizing agent in the group must look at the standard reduction potential and identify it relative to Fluorine.
• To Identify the strongest reducing agent in the group must look at the non-metal anions- I- for this example.

Calculating Cell Potential and Writing Cell Notation - Additional Examples

• Step 1: Ensure it is a voltaic or galvanic cell so the overall cell potential is positive.

Example 1: Iron and Nickel

• Given half-reactions:
  • $Fe^{3+} + e^- \rightarrow Fe^{2+}$, $E° = +0.77 V$
  • $Ni^{2+} + 2e^- \rightarrow Ni$, $E° = -0.23 V$
• Reverse the second reaction to make nickel the anode:
  • $Ni \rightarrow Ni^{2+} + 2e^-$, $E° = +0.23 V$
• Multiply the first half-reaction by 2 to balance the electrons:
  • $2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$, $E° = +0.77 V$
• Net reaction: $2Fe^{3+}(aq) + Ni(s) \rightarrow Ni^{2+}(aq) + 2Fe^{2+}(aq)$, $E° = 1.00 V$
• Anode: Nickel (oxidation).
• Cathode: Iron (reduction).
• Cell notation: $Ni(s) | Ni^{2+}(aq) || Fe^{3+}(aq), Fe^{2+}(aq) | Pt(s)$
  • Solid (Pt) electrode in the cathode because there are no solids.
  • The salt bridge is represented by the double line.

Example 2: Chromium and Chlorine

• Given Half-Reactions and Net Cell Potential:

  • $Cr^{3+} + 3e^- \rightarrow Cr(s)$, $E° = -0.74 V$
  • $Cl_2(g) + 2e^- \rightarrow 2Cl^-$, $E° = 1.36 V$

• The first step is to balance the half-reaction, for this example the least common multiple between the 2 half-reactions is 6. So adjust both sides of the reaction with 6 e-.

• Next identify which of the half-reaction will go on the anode, given Chromium in the example.

  • $2Cr(s) \rightarrow 2Cr^{3+} + 6e^-$, E° = +0.74 V
  • With the electrons one the right side this is considered oxidation.

• The other half must be positively charged, so it follows to the cathode side and thus the standard Cell Potential has to be reduction.

  • $3Cl_2(g) + 6e^- \rightarrow 6Cl^-$, E° = +1.36 V

• Putting it together and canceling 6 e's:

  • $2Cr + 3Cl_2(g) \rightarrow 2Cr^{3+} + 6Cl^-$, E° = +2.10 V

• So that the cell notation for it will be:

  • $2Cr || 2Cr^{3+} |3Cl_2(g), 6Cl^- | GraPhite$

• Now you know how to write the cell notations for different types of electrochemical cells.

Balancing Redox Reactions

• Balance redox reactions under acidic and basic conditions.

Balancing Redox Reactions & Half-Reactions Under Acidic Conditions

1. Assign oxidation numbers to each atom in the equation to determine which species are oxidized and which are reduced.
2. Separate the overall reaction into two half-reactions: one for oxidation and one for reduction.
3. Balance each half-reaction separately.
   • Balance all elements except hydrogen and oxygen.
   • Balance oxygen by adding H2O molecules to the side that needs oxygen.
   • Balance hydrogen by adding H+ ions to the side that needs hydrogen.
   • Balance charge by adding electrons (e−) to the side with the greater positive charge.
     *
4. Multiply each half-reaction by an integer so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
5. Add the two half-reactions together. Ensure that electrons cancel out.
6. Simplify the equation to the lowest possible whole-number coefficients. Verify that all elements and charges are balanced.

Example( ACID): Reducing Iodide to Form I2

1. Iodide is going to turn into $I_2$.
2. Balance the iodine atoms by adding coefficients, example: 2$I^-$. Now there are two iodine atoms on left and right side.
3. Since the charges are unbalanced you need to balance them out with electrons. You will need to add the electrons to the side of a higher charge.

Balancing the Reduction reactions and half reactions Under Acidic Conditions

1. So Chlorate ions become reduced to Chloride.
   • $ClO_3 \rightarrow Cl^{-}$
2. We now need to balance the non-oxygen or the Chlorine ions. Since they are balanced we need to move onto the Oxygen ions which are 3 here, so we balance it out as: $ClO<em>3 \rightarrow Cl^{-} + 3H</em>2O`$
3. Our end goal is to add the electrons to the left side, to do so we can the Hydrogen Ions in the following way,
   • $6H^+ + ClO<em>3 \rightarrow Cl^{-} + 3H</em>2O`$
4. Charges Need to Balance with Electrons as Follows: $6H^+ + 5e^- + ClO<em>3 \rightarrow Cl^{-} + 3H</em>2O`$
5. Final Step is to Cancel Oxidation to Reduction so to get the 5 in reduction and 2 in oxidizing to be equal, we have to multiply:
   1. Oxidizing by 5 and
   2. Reduction by multiply by 2.

Balancing Acidic to Basic Conditions

1. Keep the same half reactions and number of atoms from what we just did on the Oxidation-Reduction under Acidic conditions
2. Adding Hydroxide Ions to Both Sides:
   • So our Hydrogen becomes: 24 Hydroxide Ions.
   • Adding Water Molecules:
   • $2I^- + 24H<em>2O + 3ClO</em>3^- \rightarrow Mn^{2+} = 24OH^-$
3. Now everything under OH group can be canceled.

Electrochemical Cell and Reaction Identification

• Given an electrochemical cell with a power source and a solution containing copper sulfate and hydrochloric acid, determine the reactions occurring at the anode and cathode.

• Electrons flow from the anode to the cathode.

• Anode: oxidation.

• Cathode: reduction.

• Important Reduction Potentials (for the solution):

  • $2H+ + 2e^- \rightarrow H_2$, $E° = 0 V$
  • $Cu^{2+} + 2e^- \rightarrow Cu$, $E° = +0.34 V$
  • $O<em>2 + 4H+ + 4e^- \rightarrow 2H</em>2O$, $E° = +1.23 V$
  • $Cl_2 + 2e^- \rightarrow 2Cl^-$, $E° = +1.36 V$
  • $SO<em>4^{2-} + 4H+ + 2e^- \rightarrow SO</em>2 + 2H_2O$, $E° = +0.20 V$

• In the solution, the main species that are reduced include:

  • $H+, Cu^{2+}, SO_4^{2-}$

• Chlorides do NOT want more electrons.

• Out of these the cell with the highest positive potential wants to be reduction.

• At the Anode, the following is what can be oxidized among the given species(reversed reactions):

  • H2 can NOT occur on the anode since the gas is NOT in the solution.
  • $H_2O$ can potentially oxidize to Oxygen gas.

END

• Those were all the important notes and factors to be considering whenever thinking about the electro-chemistry and electrical potential.