Electrochemical Force Practice Questions and Answers
What Is Electrochemical Force, Anyway?
Electrochemical force is just another name for the electromotive force (EMF) or cell potential of an electrochemical cell. It tells you how much voltage a redox reaction can produce.
Units are in volts (V). A positive cell potential means the reaction happens spontaneously. A negative one means you're forcing it with external energy.
This isn't theoretical nonsense. Chemists use these calculations to predict if a reaction will work before they touch a single beaker.
The Core Formula You Need
The standard cell potential is calculated with:
E°cell = E°cathode − E°reduction
The cathode is where reduction happens. The anode is where oxidation happens. Memorize this. Students mess this up constantly.
For the Nernst equation (when conditions aren't standard):
E = E° − (RT/nF) ln(Q)
Where:
- R = 8.314 J/(mol·K)
- T = temperature in Kelvin
- n = moles of electrons transferred
- F = Faraday constant (96,485 C/mol)
- Q = reaction quotient
At 25°C, this simplifies to: E = E° − (0.0592/n) log(Q)
Practice Questions and Detailed Answers
Question 1: Calculating Standard Cell Potential
Problem: Calculate E°cell for a galvanic cell with the following half-reactions:
Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V)
Zn²⁺ + 2e⁻ → Zn (E° = −0.76 V)
Answer:
Identify the cathode and anode. Copper has the more positive potential, so it gets reduced. Zinc gets oxidized.
E°cell = E°cathode − E°anode
E°cell = (+0.34 V) − (−0.76 V)
E°cell = +1.10 V
Positive means this reaction runs spontaneously as written.
Question 2: Writing the Cell Reaction
Problem: A cell is constructed with Ag⁺/Ag (E° = +0.80 V) and Fe³⁺/Fe²⁺ (E° = +0.77 V). Write the balanced overall reaction.
Answer:
Silver has the higher reduction potential. It gets reduced. Iron gets oxidized from Fe²⁺ to Fe³⁺.
Oxidation half-reaction: Fe²⁺ → Fe³⁺ + e⁻
Reduction half-reaction: Ag⁺ + e⁻ → Ag
Multiply iron by 1 (already has 1 electron). Combine:
Ag⁺ + Fe²⁺ → Ag + Fe³⁺
E°cell = 0.80 − 0.77 = +0.03 V
This is a weak cell. The voltage is barely positive.
Question 3: Using the Nernst Equation
Problem: Calculate the cell potential at 298 K for Zn|Zn²⁺ (1.0 M) || Cu²⁺ (0.010 M)|Cu
Standard potentials: Zn²⁺/Zn = −0.76 V, Cu²⁺/Cu = +0.34 V
Answer:
First, standard cell potential: E° = 0.34 − (−0.76) = +1.10 V
Overall reaction: Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s)
Q = [Zn²⁺]/[Cu²⁺] = 1.0/0.010 = 100
n = 2 electrons
E = 1.10 − (0.0592/2) log(100)
E = 1.10 − (0.0296)(2)
E = 1.10 − 0.0592
E = 1.04 V
The lower copper concentration pushes the reaction slightly toward products, but not by much.
Question 4: Predicting Spontaneity
Problem: Will the reaction 2Ag⁺ + Cu → Cu²⁺ + 2Ag occur spontaneously?
E°(Ag⁺/Ag) = +0.80 V
E°(Cu²⁺/Cu) = +0.34 V
Answer:
Write the half-reactions as written:
Ag⁺ + e⁻ → Ag (reduction)
Cu → Cu²⁺ + 2e⁻ (oxidation)
E°cell = E°reduction − E°oxidation
E°cell = (+0.80) − (+0.34) = +0.46 V
Yes, it's spontaneous. The positive value confirms the reaction proceeds as written.
Question 5: Calculating Gibbs Free Energy
Problem: Calculate ΔG° for the reaction: MnO₂ + 4H⁺ + 2Fe²⁺ → Mn²⁺ + 2Fe³⁺ + 2H₂O
E°cell = +0.46 V (given)
Answer:
Use: ΔG° = −nFE°
n = 2 electrons (from the balanced equation)
F = 96,485 C/mol
ΔG° = −(2)(96,485)(0.46)
ΔG° = −88,766 J/mol
ΔG° = −88.8 kJ/mol
Negative ΔG° confirms spontaneity. This matches the positive E°cell.
Quick Reference: Common Standard Potentials
| Half-Reaction | E° (V) |
|---|---|
| Li⁺ + e⁻ → Li | −3.04 |
| Mg²⁺ + 2e⁻ → Mg | −2.37 |
| Al³⁺ + 3e⁻ → Al | −1.66 |
| Zn²⁺ + 2e⁻ → Zn | −0.76 |
| Fe²⁺ + 2e⁻ → Fe | −0.44 |
| 2H⁺ + 2e⁻ → H₂ | 0.00 |
| Cu²⁺ + 2e⁻ → Cu | +0.34 |
| I₂ + 2e⁻ → 2I⁻ | +0.54 |
| Ag⁺ + e⁻ → Ag | +0.80 |
| Br₂ + 2e⁻ → 2Br⁻ | +1.07 |
| Cl₂ + 2e⁻ → 2Cl⁻ | +1.36 |
| Au³⁺ + 3e⁻ → Au | +1.50 |
Where Students Actually Screw Up
These mistakes show up constantly in exams:
- Reversing the sign on the anode. You subtract E°anode, not add it. The sign is already baked in.
- Getting n wrong. Count the actual electrons in the balanced equation, not just one half-reaction.
- Forgetting to balance charges. If your equation isn't charge-balanced, the whole thing is wrong.
- Confusing Q and K. Q is for non-equilibrium conditions. K is for when the reaction has reached equilibrium (E = 0).
- Memorizing without understanding. You need to know why E°cell = E°cathode − E°anode, not just parrot the formula.
How to Actually Solve These Problems
Step-by-step approach:
- Write both half-reactions exactly as given
- Find the standard potentials for each half-reaction
- Identify cathode and anode — higher E° is reduction (cathode)
- Balance the electrons — multiply half-reactions so electrons match
- Calculate E°cell = E°cathode − E°anode
- Check your work — a positive E° means spontaneous
For Nernst equation problems:
- Write the balanced overall reaction
- Identify n (total electrons transferred)
- Calculate Q from concentrations
- Plug into Nernst equation
- Solve for E
Wrapping This Up
Electrochemical force calculations follow a small set of rules. Get the formulas down, practice balancing half-reactions, and double-check your signs.
Work through the practice questions above until you can do them without checking the answers. That's when you've actually learned this.