Chapter 16:  Oxidation-Reduction (Redox) Reactions and Electrochemistry

 

Section 16-1: Oxidation and Reduction Half-Reactions

Section 16-2: Using Oxidation Numbers to Identify Oxidizing and Reducing Agents

Section 16-3: Balancing Redox Reactions in Aqueous Acidic and Basic Solutions

Section 16-4: Using a Standard Reduction Potentials Table to Compare Strengths of Oxidizing and Reducing Agents

Section 16-5: Calculating E°cell and ΔG° to Determine Spontaneity of Redox Reactions

Section 16-6: Redox Reactions of Metals, Acids, and Halogens

Section 16-7: Experiment - Constructing a Galvanic Cell

Section 16-8: Nonspontaneous Redox Reactions and Electrolysis

Section 16-9:  Calculating the Equilibrium Constant from E°cell

Section 16-10:  The Nernst Equation - Effect of Concentration

Chapter 16 Practice Exercises and Review Quizzes

 

 

 

Section 16-1:  Oxidation and Reduction Half-Reactions

 

During the reaction Zn + Pb²⁺→ Zn²⁺ + Pb, each neutral Zn atom loses two electrons to become a Zn²⁺ cation.  These two electrons are gained by a Pb²⁺ cation, which becomes a neutral Pb atom.  The reaction can be divided into two half-reactions that show separately the loss of electrons (known as oxidation) and the gain of electrons (known as reduction). 

 

The oxidation half-reaction always shows the electrons lost on the right side of the equation:

 

oxidation:  Zn → Zn²⁺ + 2e^-

 

The reduction half-reaction always shows the electrons gained on the left side of the equation:

 

reduction:  Pb²⁺ + 2e^- → Pb

 

The overall reaction representing the sum of the oxidation and reduction half-reactions is known as an oxidation-reduction or redox reaction.  Thus, a redox reaction involves the transfer of electrons.

 

 

Section 16-2:  Using Oxidation Numbers to Identify Oxidizing and Reducing Agents

 

The reactant in the oxidation half-reaction that loses electrons is known as the reducing agent.  The reactant in the reduction half-reaction that gains electrons is known as the oxidizing agent.  In some redox reactions, it is not readily apparent which reactant is the oxidizing agent and which reactant is the reducing agent. 

 

Assigning oxidation numbers (also known as oxidation states) to each atom in the reactants and products can help us identify the oxidizing and reducing agents.  Oxidation numbers can be integers or fractions.  To assign oxidation numbers, use the following rules:

 

Rule 1:  For a given reactant or product, the sum of the oxidation numbers for all atoms is equal to the overall charge in the chemical formula.

 

Rule 2:  For neutral monatomic and polyatomic elements, the oxidation number of each atom is zero.  For example, the oxidation numbers of the iron atom in Fe, each nitrogen atom in N₂, and each sulfur atom in S₈ are all zero.

 

Rule 3:  For monatomic ions, the oxidation number is equal to the charge on the ion.  For example, the oxidation number of aluminum in Al³⁺ is +3.

 

Rule 4:  For polyatomic ions containing only one element, the oxidation number of each atom is equal to the charge divided by the number of atoms.  For example, the oxidation number of each mercury atom in Hg₂²⁺ is +1 (+2 divided by 2).

 

Rule 5:  In neutral compounds or polyatomic ions containing more than one element, oxidation numbers are assigned to each atom as follows:

 

a. Group 1 metals (does not include hydrogen) = +1

b. Group 2 metals = +2

c. H = +1

d. F = -1

e. O = -2

f. Group 17 elements = -1

g. Group 16 elements = -2

h. Group 15 elements = -3

i.  The oxidation number of elements not listed above can be determined using a combination of known oxidation numbers and Rule 1.  For example, we know from Rule 5e that the oxidation number of each oxygen atom in Cr₂O₇^2- is -2.  Since Rule 1 specifies that the sum of the oxidation numbers for all atoms should equal the overall charge of -2, we can solve the following equation to determine the oxidation number of each chromium atom:

 

2(Cr) + 7(-2) = -2

Cr = +6

 

j. When rules conflict, the rule higher on this list (closer to top of page) takes priority.  For example, SF₂ presents a conflict because Rule 5d specifies that F = -1 and Rule 5g specifies that S = -2, which does not conform to Rule 1 specifying that the sum of the oxidation numbers must equal the overall charge of zero.  Since Rule 5d specifying that F = -1 takes priority over Rule 5g, we can solve the following equation to determine the oxidation number of the sulfur atom:

 

S + 2(-1) = 0

S = +2

 

Sample Exercise 16A:

 

Determine all oxidation numbers in:

 

a. P₄

b. S^2-

c. CH₂O

d. BrO₃^-

e. NaO₂

  

Solution:

 

a. P = 0 (Rule 2)

 

b. S = -2 (Rule 3)

 

c. H = +1 (Rule 5c) and O = -2 (Rule 5e)

C + 2(+1) + (-2) = 0

C = 0

 

d. Conflict:  Rule 5e (O = -2) takes priority over Rule 5f

Br + 3(-2) = -1

Br = +5

 

e. Conflict:  Rule 5a (Na = +1) takes priority over Rule 5e

(+1) + 2(O) = 0

O = -1/2

 

 

Once we have assigned oxidation numbers to every atom in the reactants and products, we can determine which reactant is the reducing agent and which reactant is the oxidizing agent as follows:

 

reducing agent (loses electrons) = reactant containing an element that undergoes an increase in oxidation number (becomes more positive or less negative)

 

oxidizing agent (gains electrons) = reactant containing an element that undergoes a decrease in oxidation number (becomes less positive or more negative)

 

Reactions during which no elements undergo a change in oxidation number are not redox reactions. 

 

Sample Exercise 16B:

 

Determine all oxidation numbers in the following unbalanced reactions, and then decide which of the two is a redox reaction.  For the redox reaction, identify the reducing agent and the oxidizing agent. 

 

a. Na₂SO₄ + BaCl₂ → BaSO₄ + NaCl

b. VO²⁺ + F₂ → VO₂⁺ + F^- 

 

Solution:

 

Using oxidation number rules in order of priority:

 

a. Na₂SO₄ (Na = +1, O = -2, S = +6) + BaCl₂ (Ba = +2, Cl = -1)→ BaSO₄ (Ba = +2, O = -2, S = +6) + NaCl (Na = +1, Cl = -1)

 

Since no elements undergo a change in oxidation number, this is not a redox reaction.

 

b. VO²⁺ (O = -2, V = +4) + F₂ (F = 0) → VO₂⁺ (O = -2, V = +5) + F^- (F = -1)

 

Since oxidation numbers change, this is a redox reaction.  VO²⁺ contains the element V that undergoes an increase in oxidation number from +4 to +5, so VO²⁺ is the reducing agent.  F₂ contains the element F that undergoes a decrease in oxidation number from 0 to -1, so F₂ is the oxidizing agent.

 

 

Section 16-3:  Balancing Redox Reactions in Aqueous Acidic and Basic Solutions

 

Redox reactions that occur in aqueous acidic solutions can be balanced by not only changing coefficients, but also by adding hydrogen ions (H⁺) and water molecules into the reaction.  The steps required will be demonstrated in order while balancing the following equation using the smallest possible whole-number coefficients:

 

H₂C₂O₄ + NO₃^- → CO₂ + NO

 

Step 1:  Determine all oxidation numbers in the unbalanced equation, then identify the reducing agent and the oxidizing agent.

 

H₂C₂O₄ (H = +1, O = -2, C = +3) + NO₃^- (O = -2, N = +5) → CO₂ (O = -2, C = +4) + NO (O = -2, N = +2)

 

H₂C₂O₄ contains the element C that undergoes an increase in oxidation number from +3 to +4, so H₂C₂O₄ is the reducing agent.  NO₃^- contains the element N that undergoes a decrease in oxidation number from +5 to +2, so NO₃^1- is the oxidizing agent.

 

Step 2:  Write the unbalanced oxidation half-reaction and the unbalanced reduction half-reaction.  The reducing agent is always the reactant on the left side of the oxidation half-reaction, and the oxidizing agent is always the reactant on the left side of the reduction half-reaction.  The product on the right side of each half-reaction must contain the same element that underwent a change in oxidation number as the reactant in that same half-reaction:

 

oxidation:  H₂C₂O₄ → CO₂

reduction:  NO₃^- → NO

 

Step 3:  Change coefficients as necessary to balance all elements except oxygen and hydrogen:

 

oxidation:  H₂C₂O₄ → 2 CO₂

reduction:  NO₃^- → NO

 

Step 4:  Balance oxygen by adding H₂O molecules as necessary to the deficient side:

 

oxidation:  H₂C₂O₄ → 2 CO₂

reduction:  NO₃^- → NO + 2 H₂O

 

Step 5:  Balance hydrogen by adding H⁺ ions as necessary to the deficient side:

 

oxidation:  H₂C₂O₄ → 2 CO₂ + 2 H⁺

reduction:  NO₃^- + 4 H⁺ → NO + 2 H₂O

 

Step 6:  Add enough electrons to the right side of the oxidation half-reaction and the left side of the reduction half-reaction to ensure that the total charge is equal on both sides of each half-reaction.  The total charge on each side of a half-reaction can be determined by finding the sum of (coefficient x charge) for all reactants or products on that side of the equation:

 

oxidation:  H₂C₂O₄ → 2 CO₂ + 2 H⁺ + 2 e^- (total charge on left = 1 x 0 = 0, total charge on right = 2 x 0 + 2 x 1 + 2 x -1 = 0)

reduction:  NO₃^- + 4 H⁺ + 3 e^- → NO + 2 H₂O (total charge on left = 1 x -1 + 4 x 1 + 3 x -1 = 0, total charge on right = 1 x 0 + 2 x 0 = 0)

 

Step 7:  Determine the least common multiple (LCM) of the electron coefficients in each half-reaction.  Multiply all the coefficients in each half-reaction by an integer that will give both half-reactions an electron coefficient equal to the LCM:

 

LCM of electron coefficients 2 and 3 = 6

 

oxidation:  (H₂C₂O₄ → 2 CO₂ + 2 H⁺ + 2 e^-) x 3

= 3 H₂C₂O₄ → 6 CO₂ + 6 H⁺ + 6 e^- 

 

reduction:  (NO₃^- + 4 H⁺ + 3 e^- → NO + 2 H₂O) x 2

= 2 NO₃^- + 8 H⁺ + 6 e^- → 2 NO + 4 H₂O

 

Step 8:  Add the two half-reactions to obtain the overall balanced equation, which should no longer contain electrons and should have the same total charge on both sides:

 

3 H₂C₂O₄ + 2 NO₃^- + 8 H⁺ + 6e^- → 6 CO₂ + 6 H⁺ + 6 e^- + 2 NO + 4 H₂O

overall balanced equation: 3 H₂C₂O₄ + 2 NO₃^- + 2 H⁺ → 6 CO₂ + 2 NO + 4 H₂O

(total charge on left = 3 x 0 + 2 x -1 + 2 x 1 = 0, total charge on right = 6 x 0 + 2 x 0 + 4 x 0 = 0)

 

Sample Exercise 16C:

 

Determine all oxidation numbers, identify the reducing agent and oxidizing agent, and balance the following equation that occurs in aqueous acidic solution using the smallest possible whole-number coefficients:

 

Cr₂O₇^2- + I₂ → Cr³⁺ + IO₃^-

Solution:

 

Step 1: 

Cr₂O₇^2- (O = -2, Cr = +6) + I₂ (I = 0) → Cr³⁺ (Cr = +3) + IO₃^- (O = -2, I = +5)

reducing agent = I₂ (oxidation number of I increases from 0 to +5)

oxidizing agent = Cr₂O₇^2- (oxidation number of Cr decreases from +6 to +3)

 

Step 2:

oxidation:  I₂ → IO₃^-

reduction:  Cr₂O₇^2- → Cr³⁺

 

Step 3:

oxidation:  I₂ → 2 IO₃^-

reduction:  Cr₂O₇^2- → 2 Cr³⁺

 

Step 4:

oxidation:  I₂ + 6 H₂O → 2 IO₃^-

reduction:  Cr₂O₇^2- → 2 Cr³⁺ + 7 H₂O

 

Step 5:

oxidation:  I₂ + 6 H₂O → 2 IO₃^- + 12 H⁺

reduction:  Cr₂O₇^2- + 14 H⁺ → 2 Cr³⁺ + 7 H₂O

 

Step 6:

oxidation:  I₂ + 6 H₂O → 2 IO₃^- + 12 H⁺ + 10 e^-

reduction:  Cr₂O₇^2- + 14 H⁺ + 6 e^- → 2 Cr³⁺ + 7 H₂O

 

Step 7:  LCM of electron coefficients 10 and 6 = 30

oxidation:  (I₂ + 6 H₂O → 2 IO₃^- + 12 H⁺ + 10 e^-) x 3

= 3 I₂ + 18 H₂O → 6 IO₃^- + 36 H⁺ + 30 e^-

 

reduction:  (Cr₂O₇^2- + 14 H⁺ + 6 e^- → 2 Cr³⁺ + 7 H₂O) x 5

= 5 Cr₂O₇^2- + 70 H⁺ + 30 e^- → 10 Cr³⁺ + 35 H₂O

 

Step 8:

3 I₂ + 18 H₂O + 5 Cr₂O₇^2- + 70 H⁺ + 30 e^- → 6 IO₃^- + 36 H⁺ + 30 e^- + 10 Cr³⁺ + 35 H₂O

overall balanced equation: 3 I₂ + 5 Cr₂O₇^2- + 34 H⁺ → 6 IO₃^- + 10 Cr³⁺ + 17 H₂O

(total charge on left = 3 x 0 + 5 x -2 + 34 x 1 = +24, total charge on right = 6 x -1 + 10 x 3 + 17 x 0 = +24)

 

To balance redox reactions that do not involve oxygen or hydrogen, omit Step 4 and Step 5 from the method above.

 

To balance a redox reaction that occurs in aqueous basic solution, first follow Steps 1 through 8 above as if the reaction is acidic and then continue as follows:

 

Step 9:  If y = coefficient of H⁺ after Step 8 is complete, add y OH^- ions to both sides of the equation. 

 

Step 10:  On the side of the equation with both H⁺ and OH^-, combine y H⁺ with y OH^- to create y H₂O and then simplify to obtain the final balanced equation with H₂O on only one side of the equation.

 

 Sample Exercise 16D:

Determine all oxidation numbers, identify the reducing agent and oxidizing agent, and balance the following equation that occurs in aqueous basic solution using the smallest possible whole-number coefficients:

 

MnO₂ + As → Mn₂O₃ + AsO₂^-

 

Solution:

 

Steps 1 through 8: 

MnO₂ (O = -2, Mn = +4) + As (As = 0) → Mn₂O₃(O = -2, Mn = +3) + AsO₂^- (O = -2, As = +3)

reducing agent = As (oxidation number of As increases from 0 to +3)

oxidizing agent = MnO₂ (oxidation number of Mn decreases from +4 to +3)

 

oxidation:  (As + 2 H₂O → AsO₂^- + 4 H⁺ + 3 e^-) x 2

reduction:  (2 MnO₂ + 2 H⁺ + 2 e^-→ Mn₂O₃ + H₂O) x 3

overall balanced equation (acidic):  2 As + 6 MnO₂ + H₂O → 2 AsO₂^- + 3 Mn₂O₃ + 2 H⁺

 

Step 9:

2 As + 6 MnO₂ + H₂O + 2 OH^-→ 2 AsO₂^- + 3 Mn₂O₃ + 2 H⁺ + 2 OH^-

 

Step 10:

2 As + 6 MnO₂ + H₂O + 2 OH^-→ 2 AsO₂^- + 3 Mn₂O₃ + 2 H₂O

overall balanced equation (basic):  2 As + 6 MnO₂ + 2 OH^-→ 2 AsO₂^- + 3 Mn₂O₃ + H₂O

 

 

 

Section 16-4:  Using a Standard Reduction Potentials Table to Compare Strengths of Oxidizing and Reducing Agents

The table of standard reduction potentials (E°_red) in the unit volts (V) shown below can be used to compare the relative tendency of reduction half-reactions to occur under standard conditions (all aqueous reactants and products having a concentration of 1 M and all gaseous reactants and products having a partial pressure of 1 atm):

 

 

When E°_red is more positive or less negative, the reduction half-reaction will have a greater tendency to occur.  This means that the reactant in the reduction half-reaction will have a greater tendency to gain electrons, making it a stronger oxidizing agent.

 

 

Sample Exercise 16E:

Rank the aqueous cations Ag⁺, Au³⁺, Co²⁺, and Sn²⁺ from weakest to strongest oxidizing agent under standard conditions. Justify your answer using a table of standard reduction potentials.

 

Solution:

 

The cations should be ranked from most negative to most positive E°_red:

 

weakest oxidizing agent = Co²⁺ (-0.28 V) < Sn²⁺ (-0.14 V) < Ag⁺ (+0.80 V) < Au³⁺ (+1.50 V) = strongest oxidizing agent

 

 

Any reduction half-reaction on the table above can be reversed to yield an oxidation half-reaction with a standard oxidation potential, E°_ox, equal in magnitude but opposite in sign to E°_red for the reduction half-reaction.  For example:

 

2 F^- (aq) → F₂ (g) + 2 e^-     E°_ox = -2.87 V

Li (s) →  Li⁺ (aq) + 1 e^-     E°_ox = +3.05 V

 

When E°_ox is more positive or less negative, the oxidation half-reaction will have a greater tendency to occur.  This means that the reactant in the oxidation half-reaction will have a greater tendency to lose electrons, making it a stronger reducing agent. 

 

Sample Exercise 16F:

 

Rank the aqueous cations Co²⁺, Fe²⁺, and Sn²⁺ from weakest to strongest reducing agent under standard conditions. Justify your answer using a table of standard reduction potentials.

 

Solution:

 

First, we reverse the appropriate half-reactions from the table above:

 

Co²⁺ (aq) → Co³⁺ (aq) + 1 e^-     E°_ox = -1.82 V

Fe²⁺ (aq) → Fe³⁺ (aq) + 1 e^-     E°_ox = -0.77 V

Sn²⁺ (aq) → Sn⁴⁺ (aq) + 2 e^-     E°_ox = -0.15 V

 

We then rank the cations from most negative to least negative E°_ox:

 

weakest reducing agent = Co²⁺ < Fe²⁺ < Sn²⁺ = strongest reducing agent

 

 

Section 16-5:  Calculating E°_cell and ΔG° to Determine Spontaneity of Redox Reactions

The standard cell potential (or standard cell voltage), E°_cell, for a redox reaction can be calculated by adding the standard oxidation potential for the oxidation half-reaction to the standard reduction potential for the reduction half-reaction:

 

E°_cell = E°_ox + E°_red

 

If E°_cell is positive, the reaction will be spontaneous under standard conditions.  If E°_cell is negative, the reaction will be nonspontaneous under standard conditions.

 

ΔG° for a redox reaction can be calculated from E°_cell using the following equation:

 

ΔG° = -nFE°_cell

 

n represents the number of electrons transferred in the balanced equation and is equal to the LCM of the electron coefficients in the oxidation and reduction half-reactions prior to Step 7 of the balancing process.  F is known as Faraday's Constant and represents the electric charge in the unit coulombs (C) per mole of electrons.  F has a value of 96,500 C/mol.  Given that 1 V = 1 J/C, we will replace the E°_cell unit V with J/C when calculating ΔG° from E°_cell: 

 

Sample Exercise 16G:

Calculate E°_cell and ΔG° for the reaction Cr (s) + Sn⁴⁺ (aq) → Cr³⁺ (aq) + Sn²⁺ (aq), determine if the reaction is spontaneous or nonspontaneous under standard conditions, and then balance the equation.

 

Solution:

 

The half-reactions prior to Step 7 of the balancing process are:

 

oxidation:  Cr (s) → Cr³⁺ (aq) + 3 e^-     E°_ox = +0.74 V

reduction:  Sn⁴⁺ (aq) + 2 e^- → Sn²⁺ (aq)     E°_red = +0.15 V

 

Thus, E°_cell = E°_ox + E°_red = 0.74 V + 0.15 V = 0.89 V.  Note that multiplying the coefficients in each half-reaction during Step 7 of the balancing process will not change the values of E°_ox or E°_red.  ΔG° can be calculated from E°_cell as follows:

 

n = LCM of electron coefficients 3 and 2 = 6

 

Since E°_cell > 0 and ΔG° < 0, the reaction is spontaneous under standard conditions.  The remaining steps in the balancing process are:

 

oxidation:  (Cr (s) → Cr³⁺ (aq) + 3 e^-) x 2

reduction:  (Sn⁴⁺ (aq) + 2 e^- → Sn²⁺ (aq)) x 3

 

overall balanced equation:  2 Cr (s) + 3 Sn⁴⁺ (aq) → 2 Cr³⁺ (aq) + 3 Sn²⁺ (aq)

  

 

Section 16-6:  Redox Reactions of Metals, Acids, and Halogens

 

(a) solid metal + aqueous metal nitrate or aqueous metal sulfate

 

Aqueous nitrate and sulfate ions are generally spectator ions during redox reactions occurring in non-acidic solutions that lack a significant concentration of H⁺ ions.  As such, when a solid metal reacts with an aqueous metal nitrate or metal sulfate solution, the half-reactions follow the pattern:

 

oxidation:  M₁ (s) → M₁^m+ (aq) + m e^-

reduction:  M₂ⁿ⁺ (aq) + n e^- → M₂ (s)

 

Therefore, the unbalanced net ionic equation will be:

 

M₁ (s) + M₂ⁿ⁺ (aq) → M₁^m+ (aq) + M₂ (s)

 

(b) solid metal + aqueous hydrochloric acid

 

HCl ionizes completely into H⁺ and Cl^- in water.  When a solid metal reacts an aqueous hydrochloric acid solution, the chloride ion will be a spectator ion.  As such, the half-reactions follow the pattern:

 

oxidation:  M (s) → M^m+ (aq) + m e^-

reduction:  2 H⁺ (aq) + 2 e^- → H₂ (g)

 

 Therefore, the unbalanced net ionic equation will be:

 

M (s) + H⁺ (aq) → M^m+ (aq) + H₂ (g)

 

(c) solid metal + aqueous nitric acid

 

HNO₃ ionizes completely into H⁺ and NO₃^- in water.  When a solid metal reacts with an aqueous nitric acid solution, since the nitrate ion is a strong oxidizing agent in acidic solutions, the half-reactions follow the pattern:

 

oxidation:  M (s) → M^m+ (aq) + m e^-

reduction:  NO₃^- (aq) + 4 H⁺ (aq) + 3 e^- → NO (g) + 2 H₂O (l)

 

Therefore, the unbalanced net ionic equation will be:

 

M (s) + NO₃^- (aq) + H⁺ (aq) → M^m+ (aq) + NO (g) + H₂O (l)

 

(d) diatomic halogen + aqueous metal halide

 

When a diatomic halogen reacts with an aqueous metal halide solution, the metal ion will be a spectator ion.  As such, the half-reactions will follow the pattern:

 

oxidation:  2 X^- (aq) → X₂ (s, l, or g) + 2 e^-

reduction:  Y₂ (s, l, or g) + 2 e^- → 2 Y^- (aq)

 

Therefore, the unbalanced net ionic equation will be:

 

X^- (aq) + Y₂ (s, l, or g) → X₂ (s, l, or g) + Y^- (aq)

 

Sample Exercise 16H:

 

Which reaction in each pair below will be spontaneous under standard conditions?  For each spontaneous reaction, calculate E°_cell and ΔG°, then balance the equation.

 

1a. NiSO₄ (aq) + solid manganese metal or 1b. NiSO₄ (aq) + solid lead metal?

 

2a. aqueous hydrochloric acid + solid magnesium metal or 2b. aqueous hydrochloric acid + solid silver metal?

 

3a. aqueous nitric acid + solid copper metal or 3b. aqueous nitric acid + solid gold metal?

 

4a. liquid bromine + aqueous potassium iodide or 4b. solid iodine + aqueous sodium bromide? 

 

Solution:

1a. SO₄^2- = spectator ion

oxidation:  Mn (s) → Mn²⁺ (aq) + 2 e^-     E°_ox = +1.18 V

reduction:  Ni²⁺ (aq) + 2 e^- → Ni (s)     E°_red = -0.25 V

 

E°_cell = 1.18 V + (-0.25 V) = 0.93 V > 0 = spontaneous

 

 

balanced equation: Mn (s) + Ni²⁺ (aq) → Mn²⁺ (aq) + Ni (s)

 

 

1b. SO₄^2- = spectator ion

oxidation:  Pb (s) → Pb²⁺ (aq) + 2 e^-     E°_ox = +0.13 V

reduction:  Ni²⁺ (aq) + 2 e^- → Ni (s)     E°_red = -0.25 V

 

E°_cell = 0.13 V + (-0.25 V) < 0 = nonspontaneous

 

2a. Cl^- = spectator ion

oxidation:  Mg (s) → Mg²⁺ (aq) + 2 e^-     E°_ox = +2.37 V

reduction:  2 H⁺ (aq) + 2 e^- → H₂ (g)     E°_red = 0 V

 

E°_cell = 2.37 V + 0 V = 2.37 V > 0 = spontaneous

 

 

balanced equation: Mg (s) + 2 H⁺ (aq) → Mg²⁺ (aq) + H₂ (g)

 

2b. Cl^- = spectator ion

oxidation:  Ag (s) → Ag⁺ (aq) + 1 e^-     E°_ox = -0.80 V

reduction:  2 H⁺ (aq) + 2 e^- → H₂ (g)     E°_red = 0 V

 

E°_cell = (-0.80 V) + 0 V < 0 = nonspontaneous

 

3a.

oxidation:  Cu (s) → Cu²⁺ (aq) + 2 e^-     E°_ox = -0.34 V

reduction:  NO₃^- (aq) + 4 H⁺ (aq) + 3 e^- → NO (g) + 2 H₂O (l)     E°_red = +0.96 V

 

E°_cell = (-0.34 V) + 0.96 = 0.62 V > 0 = spontaneous

 

 

balanced equation:  3 Cu (s) + 2 NO₃^- (aq) + 8 H⁺ (aq) → 3 Cu²⁺ (aq) + 2 NO (g) + 4 H₂O (l)

 

 

3b.

oxidation:  Au (s) → Au³⁺ (aq) + 3 e^-     E°_ox = -1.50 V

reduction:  NO₃^- (aq) + 4 H⁺ (aq) + 3 e^- → NO (g) + 2 H₂O (l)     E°_red = +0.96 V

 

E°_cell = (-1.50 V) + 0.96 < 0 = nonspontaneous

 

4a. K⁺ = spectator ion

oxidation:  2 I^- (aq) → I₂ (s) + 2 e^-     E°_ox = -0.53 V

reduction:  Br₂ (l) + 2 e^- → 2 Br^- (aq)     E°_red = 1.07 V

 

E°_cell = (-0.53 V) + 1.07 = 0.54 V > 0 = spontaneous

 

 

balanced equation:  2 I^- (aq) + Br₂ (l) → I₂ (s) + 2 Br^- (aq)

 

 

4b. Na⁺ = spectator ion

oxidation:  2 Br^- (aq) → Br₂ (l) + 2 e^-     E°_ox = -1.07 V

reduction:  I₂ (s) + 2 e^- → 2 I^- (aq)     E°_red = +0.53 V

 

E°_cell = (-1.07 V) + 0.53 < 0 = nonspontaneous

 

Section 16-7:  Experiment – Constructing a Galvanic Cell

A galvanic cell (or voltaic cell) utilizes a spontaneous redox reaction to provide electrical energy.  The oxidation and reduction half-reactions are separated into two different compartments.  As such, the electrons transferred must flow from the location of oxidation to the location of reduction through a metal wire.  This spontaneous movement of electrons provides electrical energy.

 

A galvanic cell can be constructed using standard laboratory chemicals and equipment as follows:

 

1. Immerse one end of a metal strip in a beaker containing aqueous ions of that same metal.  Repeat in a second beaker using a different metal.  Note that the anions in each beaker will be spectator ions.

 

2. Connect the non-immersed ends of the two metal strips using a metal wire.  A voltmeter can be attached along the wire to read the voltage as the galvanic cell operates.  Note that if 1 M aqueous solutions are used in the beakers, then the reading on the voltmeter will be equal to E°_cell for the spontaneous reaction.

 

3. Connect the aqueous solutions in the two beakers using a salt bridge that can be constructed using a U-shaped glass tube containing an aqueous ionic salt.  Note that the ions in the salt bridge will not be part of the spontaneous reaction.  The purpose of the salt bridge is to keep the solutions in each beaker electrically neutral, which means that the total positive charge equals the total negative charge.  Negatively-charge anions will flow out from the salt bridge into the beaker where oxidation is occurring to counteract the positively-charged metal cations being created there.  Positively-charged cations will flow out from the salt bridge into the beaker where reduction is occurring to replace the positively-charged metal cations being consumed there.

 

A sketch of a galvanic cell in operation is shown below:    

 

 

Key aspects of the galvanic cell are:

 

1. The metal electrode that undergoes oxidation (M₁) is known as the anode and is typically labeled (-).  The anode will lose mass as the reaction proceeds.  The metal electrode that is the product of reduction (M₂) is known as the cathode and is typically labeled (+).  The cathode will gain mass as the reaction proceeds. 

 

2. Electrons flow through the wire from anode to cathode, or from (-) to (+). 

 

3. Anions flow out of the salt bridge toward the anode.  Cations flow out of the salt bridge toward the cathode.       

 

Sample Exercise 16I:

A galvanic cell was constructed using a strip of aluminum metal and a strip of zinc metal, a 1 M solution of Al(NO₃)₃ and a 1 M solution of ZnSO₄, and an aqueous solution of KNO₃ in the salt bridge.  For the spontaneous reaction that occurred, calculate E°_cell and ΔG°, then balance the equation.  Also sketch the galvanic cell.

 

Solution:

 

We must choose the correct combination of half-reactions to obtain a spontaneous reaction (nitrate and sulfate = spectator ions):

 

oxidation:  Al (s) → Al³⁺ (aq) + 3 e^-     E°_ox = +1.66 V

reduction:  Zn²⁺ (aq) + 2 e^- → Zn (s)     E°_red = -0.76 V

 

E°_cell = 1.66 V + (-0.76 V) = 0.90 V > 0 = spontaneous

 

 

balanced equation: 2 Al (s) + 3 Zn²⁺ (aq) → 2 Al³⁺ (aq) + 3 Zn (s)

 

Redraw the galvanic cell sketch shown above, making the following substitutions:

 

M₁ = Al, M₁^m+ = Al³⁺

M₂ = Zn, M₂ⁿ⁺ = Zn²⁺

anions = NO₃^-, cations = K⁺

 

Section 16-8:  Nonspontaneous Redox Reactions and Electrolysis

A nonspontaneous redox reaction can still occur through a process known as electrolysis if a high enough voltage is applied.  As is the case for a galvanic cell, oxidation occurs at the anode and reduction occurs at the cathode in an electrolytic cell.  The minimum voltage required for electrolysis under standard conditions is equal to the absolute value of E°_cell for the nonspontaneous reaction:

 

electrolysis:  minimum voltage = |E°_cell|

 

Sample Exercise 16J:

Calculate the minimum voltage required to bring about the reaction Br^- (aq) + Fe³⁺ (aq) → Br₂ (l) + Fe²⁺ (aq) by electrolysis under standard conditions, then balance the equation.

  

Solution:

oxidation (at anode):  2 Br^- (aq) → Br₂ (l) + 2 e^-     E°_ox = -1.07 V

reduction (at cathode):  Fe³⁺ (aq) + 1 e^- → Fe²⁺ (aq)     E°_red = +0.77 V

 

E°_cell = (-1.07 V) + 0.77 = -0.30 V

minimum voltage = |-0.30 V| = 0.30 V

 

balanced equation:  2 Br^- (aq) + 2 Fe³⁺ (aq) → Br₂ (l) + 2 Fe²⁺ (aq)

 

If a liquid (also known as molten) ionic metal halide such as LiCl, CaBr₂, or NaI is electrolyzed, the halide anion will be oxidized at the anode to a neutral diatomic halogen while the metal cation will be reduced at the cathode to a neutral metal:

 

oxidation (at anode):  2 X ^-→ X₂ + 2 e^-

reduction (at cathode):  Mⁿ⁺ + n e^- → M

 

Sample Exercise 16K:

For the electrolysis of molten LiCl, write the half-reaction that occurs at the anode and the half-reaction that occurs at the cathode, then balance the equation.

  

Solution:

 

oxidation (at anode):  2 Cl^- → Cl₂ + 2 e^-

reduction (at cathode):  Li⁺ + 1 e^- → Li

 

balanced equation:  2 Cl^- + 2 Li⁺ → Cl₂ + 2Li

The total charge in coulombs (C) transferred during electrolysis can be found by multiplying the time of electrolysis in seconds by the current in amperes (A), where 1 A = 1 C/s:

 

(time in s) x (current in C/s) = charge in C

 

Once we know the total charge transferred in C, we can divide by F = 96,500 C/mol e^- to find the total moles of electrons transferred during electrolysis.  From the total moles of electrons transferred, we can use the mole ratio from the relevant half-reaction to find the moles of a product of interest, after which moles of product can be converted to any desired unit. 

 

Starting with time, the general strategy employed in order to determine the quantity of a product formed during electrolysis is as follows:

 

 

Starting with the quantity of product, the general strategy employed in order to determine the time of electrolysis is as follows:

 

 

We will use dimensional analysis to implement each strategy above as follows:

 

Sample Exercise 16L:

For the electrolysis of molten LiCl: 

 

a. If the electrolyis proceeds for 244 minutes using a current of 0.485 A, how many milliliters of Cl₂ gas, measured at 35°C and 711 mmHg, will be produced? 

b. Using a current of 1.8 A, how many hours must the electrolysis proceed to produce 1.3 g of Li?

  

Solution:

 

a. The relevant half-reaction is:

 

oxidation (at anode):  2 Cl^- → Cl₂ + 2 e^-

 

b. The relevant half-reaction is:

 

reduction (at cathode):  Li⁺ + 1 e^- → Li

 

 

When an aqueous solution is electrolyzed, in some cases the solvent water rather than the solute particles will be oxidized at the anode and/or reduced at the cathode according to the following half-reactions:

 

oxidation (at anode):  2 H₂O (l) → O₂ (g) + 4 H⁺ (aq) + 4 e^-     E°_ox = -1.23 V

reduction (at cathode):  2 H₂O (l) + 2 e^-→ H₂ (g) + 2 OH^- (aq)     E°_red = -0.83 V

 

For example, when MgI₂ (aq) is electrolyzed, the reduction of water shown above will occur at the cathode rather than the reduction of Mg²⁺ [Mg²⁺ (aq) + 2 e^-→ Mg (s), E°_red = -2.37 V] because E°_red is less negative for the reduction of water.  On the other hand, the oxidation of I^- [2 I^- (aq) → I₂ (s) + 2 e^-, E°_ox = -0.53 V] will occur at the anode rather than the oxidation of water shown above because E°_ox is less negative for the oxidation of I^-. 

 

Although we would generally expect the oxidation half-reaction with the least negative E°_ox to occur at the anode, one notable exception is that the production of Cl₂ (g) may be observed at the anode during the electrolysis of Cl^- (aq) despite E°_ox for the oxidation of Cl^- [2 Cl^- (aq) → Cl₂ (g) + 2 e^-, E°_ox = -1.36 V] being more negative than E°_ox for the oxidation of water.

 

The alkali metal cations Li⁺ (aq), Na⁺ (aq), and K⁺ (aq) and alkaline earth metal cations Ca²⁺ (aq) and Mg²⁺ (aq) can be considered spectator ions during electrolysis.  In addition, the anions NO₃^- (aq) and SO₄^2- (aq) can be considered spectator ions if the solution being electrolyzed is not acidic.

 

If we know the time of electrolysis and the amount of a product formed, we can calculate the current used as follows:

 

Sample Exercise 16M:

a. For the electrolysis of aqueous K₂SO₄, write the half-reaction that occurs at the anode and the half-reaction that occurs at the cathode, then balance the equation.

b. If 2.0 liters of oxygen gas, measured at 22°C and 713 torr, are produced when the electrolysis proceeds for 3.0 hours, calculate the current used in the electrolysis.

 

Solution:

 

a. K⁺ (aq) and SO₄^2- (aq) = electrolysis spectator ions:

 

oxidation (at anode):  2 H₂O (l) → O₂ (g) + 4 H⁺ (aq) + 4 e^-

reduction (at cathode):  (2 H₂O (l) + 2 e^-→ H₂ (g) + 2 OH^- (aq)) x 2

balanced equation:  2 H₂O (l) → 2 H₂ (g) + O₂ (g)

 

b. The relevant half-reaction is:

 

oxidation (at anode):  2 H₂O (l) → O₂ (g) + 4 H⁺ (aq) + 4 e^-

 

 

Section 16-9:  Calculating the Equilibrium Constant from E°_cell

Given that ΔG° = -nFE°_cell and ΔG° = -RTlnK_eq, then -nFE°_cell = -RTlnK_eq.  Dividing both sides of this equation by -nF yields the equation E°_cell = (RT/nF)lnK_eq.  If we focus only on reactions at 25°C, substituting 298 K for T as well as R = 8.31 J/mol•K and F = 96,500 C/mol yields E°_cell = (8.31 J/mol•K x 298 K)/(n x 96,500 C/mol)lnK_eq.  This can be simplified to obtain the following equation relating E°_cell and K_eq:

 

E°_cell = (0.0257 V/n)lnK_eq

 

Sample Exercise 16N:

Calculate K_c for the reaction 2 Cr (s) + 3 Zn²⁺ (aq) → 2 Cr³⁺ (aq) + 3 Zn (s) at 25°C. 

 

Solution:

 

oxidation:  Cr (s) → Cr³⁺ (aq) + 3 e^-     E°_ox = +0.74 V

reduction:  Zn²⁺ (aq) + 2 e^-→ Zn (s)     E°_red = -0.76 V

 

E°_cell = 0.74 V + (-0.76 V) = -0.02 V

 

K_eq = K_c:

-0.02 V = (0.0257 V/6)lnK_c

K_c = 9 x 10^-3

 

 

Section 16-10:  The Nernst Equation – Effect of Concentration

For a nonstandard galvanic cell constructed using at least one aqueous reactant or product with a molarity not equal to 1 M (and no gases), the following version of the Nernst equation can be used to calculate the cell potential (or cell voltage), E_cell:

 

E_cell = E°_cell – (0.0257 V/n)lnQ_c

 

The effect on E_cell of changing either reactant concentrations only or product concentrations only can be predicted as follows:

 

increase reactant concentrations only = Q_c↓ = E_cell ↑

decrease reactant concentrations only = Q_c↑ = E_cell ↓

increase product concentrations only = Q_c↑ = E_cell ↓

decrease product concentrations only = Q_c↓ = E_cell ↑

 

Sample Exercise 16O:

A galvanic cell is constructed at 25°C that utilizes the reaction Sn²⁺ (aq) + 2 Ag⁺ (aq) → Sn⁴⁺ (aq) + 2 Ag (s).

 

a. If all aqueous ions in the cell have an initial concentration of 1 M, calculate E°_cell.

b. Predict whether E_cell will increase, decrease, or remain unchanged if the initial concentrations of Sn²⁺ and Ag⁺ are 1 M, but the initial concentration of Sn⁴⁺ is changed to 2 M.

c. Calculate E_cell if the initial concentrations of all aqueous ions are 0.25 M.  

 

Solution:

 

a.

oxidation:  Sn²⁺ (aq) → Sn⁴⁺ (aq) + 2 e^-     E°_ox = -0.15 V

reduction:  Ag⁺ (aq) + 1 e^-→ Ag (s)     E°_red = +0.80 V

 

E°_cell = (-0.15 V) + 0.80 V = 0.65 V

 

b. increase product concentration only = Q_c increases = E_cell decreases

 

c.

 

 

 

Chapter 16 Practice Exercises and Review Quizzes:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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