Required Practical 8 • CH03 / CH05

RP8: Measuring EMF of Electrochemical Cells

Construct electrochemical half-cells, measure cell potentials using a high-resistance digital voltmeter, prepare potassium nitrate salt bridges, and evaluate Le Chatelier concentration shifts on cell voltage.

1. Principles of Electrochemical Cells

Required Practical 8 explores how chemical energy is converted into electrical energy in galvanic cells. When two different redox half-cells are coupled, a potential difference (electromotive force, EMF) arises from the difference in electron-releasing ability (standard electrode potential, E°) between the two couples.

The Daniell Cell Standard Benchmark Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
- Anode (Negative Electrode): Zinc undergoes oxidation → Zn(s) → Zn2+(aq) + 2e- (E° = -0.76 V).
- Cathode (Positive Electrode): Copper(II) undergoes reduction → Cu2+(aq) + 2e- → Cu(s) (E° = +0.34 V).
- Standard EMF: E°_cell = E°(cathode) - E°(anode) = +0.34 - (-0.76) = +1.10 V.

2. Half-Cell Construction & Electrode Preparation

Accurate EMF measurement requires careful physical preparation of metal electrodes and standard solutions:

  1. Cleaning Metal Strips: Clean strips of zinc, copper, and iron with emery paper or sandpaper to scour off the oxide coating (e.g. ZnO, CuO). Rinse with deionised water and dry with a paper towel.
  2. Half-Cell Solutions: Measure 50 cm3 of 1.00 mol dm^-3 ZnSO4(aq) into one beaker, and 50 cm3 of 1.00 mol dm^-3 CuSO4(aq) into a second beaker.
  3. Immerse Electrodes: Place the zinc strip into the ZnSO4 solution, and the copper strip into the CuSO4 solution.
  4. Inert Electrodes for Solutions: For half-cells consisting of two ions in the same phase (e.g. Fe3+(aq) / Fe2+(aq)), an inert platinum (Pt) electrode must be used to conduct electrons without reacting chemically.
Zinc-Copper Electrochemical Cell Setup Electrochemical Cell Apparatus (Daniell Cell) V +1.10 V (High Resistance) Zn strip (-) 1.0 M ZnSO4(aq) Cu strip (+) 1.0 M CuSO4(aq) KNO3 Salt Bridge Ions complete circuit e- flow →

3. The Salt Bridge: Function, Preparation & Rules

The salt bridge is a vital electrical link between the two separate half-cells:

Function of the Salt Bridge

  • Completes the Circuit: Allows the free movement of ions between beakers so current can flow without transferring the bulk electrolyte solutions.
  • Maintains Electrical Neutrality: As Zn2+ ions enter the left beaker, anions (NO3-) migrate out of the bridge into the zinc beaker. As Cu2+ ions discharge at the right electrode, cations (K+) migrate into the copper beaker.

Preparation & Chemical Compatibility Rules

  • Preparation: Soak a strip of clean filter paper in saturated potassium nitrate (KNO3) or potassium chloride (KCl) solution.
  • Crucial Exclusion Rule: Never use a potassium chloride (KCl) salt bridge if either half-cell contains silver (Ag+) or lead (Pb2+) ions. Insoluble AgCl or PbCl2 precipitates would crystallize inside the porous paper, blocking ion pores and breaking the electrical circuit. Always use KNO3 as the universal bridge.

4. Why a HIGH-RESISTANCE Voltmeter Must Be Used

A classic Paper 5 examination question challenges candidates on the electrical properties of the measuring instrument:

Mandate for High Resistance A standard digital voltmeter has an internal resistance > 10^6 Ω (mega-ohms).

Why High Resistance is Essential:
1. By Ohm's law (I = V / R), a massive internal resistance ensures that virtually zero electric current flows through the external circuit.
2. If current flowed, electrons would be consumed, driving the cell reaction forward. Reactant concentrations would drop, products would accumulate, and the system would shift away from equilibrium.
3. Drawing zero current measures the true maximum potential difference (electromotive force) under reversible, non-polarized equilibrium conditions.

5. Calculating Standard Cell Potential (E°_cell)

The electromotive force is calculated from standard electrode potentials listed in the data booklet:

E°_cell = E°(Right / Cathode / Reduction) - E°(Left / Anode / Oxidation)

Standard Conditions:

  • Temperature: 298 K (25 deg C)
  • Pressure: 100 kPa (1 bar) for gaseous half-cells (e.g. H2 gas in the standard hydrogen electrode)
  • Concentration: 1.00 mol dm^-3 for all aqueous ions

6. Le Chatelier Concentration Shifts on Cell EMF

When ion concentrations deviate from 1.00 mol dm^-3, the equilibrium position of the half-cell reaction shifts, altering the electrode potential and overall cell voltage:

Concentration Change Equilibrium Shift (Le Chatelier) Effect on Half-Cell Potential Effect on Overall Cell EMF
Diluting Zn2+ at Anode (e.g. from 1.0 M down to 0.10 M) Zn2+ + 2e- ⇔ Zn shifts left to replace lost ions. Zinc releases electrons more readily; E(anode) becomes more negative. Cell EMF INCREASES (e.g. rises from 1.10 V to ~1.13 V).
Diluting Cu2+ at Cathode (e.g. from 1.0 M down to 0.10 M) Cu2+ + 2e- ⇔ Cu shifts left. Copper accepts electrons less readily; E(cathode) becomes less positive. Cell EMF DECREASES (falls from 1.10 V to ~1.07 V).

7. Worked Electrochemical Cell Problem

Worked Example: Cell EMF and Spontaneous Direction

Problem: An electrochemical cell is constructed using an iron half-cell and a silver half-cell under standard conditions:

  • Fe2+(aq) + 2e- ⇔ Fe(s)    E° = -0.44 V
  • Ag+(aq) + e- ⇔ Ag(s)      E° = +0.80 V

1. Identify the negative electrode (anode) and positive electrode (cathode).
2. Write the standard conventional cell representation.
3. Calculate the standard EMF (E°_cell).
4. Write the overall spontaneous cell equation.

Step 1: Identify Electrodes

Fe2+/Fe has more negative E° (-0.44 V) → Oxidation → Anode (Negative).
Ag+/Ag has more positive E° (+0.80 V) → Reduction → Cathode (Positive).

Step 2: Conventional Cell Representation

Fe(s) | Fe2+(aq) || Ag+(aq) | Ag(s)

Step 3: Calculate Standard EMF

E°_cell = E°(cathode) - E°(anode)
E°_cell = +0.80 - (-0.44) = +1.24 V

Step 4: Overall Balanced Equation

Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s)

Final Answer: E°_cell = +1.24 V

8. Practice Exam Questions

Question 1: Why is potassium nitrate (KNO3) preferred over potassium chloride (KCl) for preparing a salt bridge in an electrochemical cell that includes a silver half-cell (Ag+/Ag)?

Show Answer & Explanation

Correct Answer: B

Chloride ions diffusing out of a KCl bridge react with Ag+ in solution to precipitate insoluble AgCl(s), clogging the filter paper and depleting free silver ions. All nitrates are completely soluble, making KNO3 non-interfering.

Question 2: What would happen to the measured potential of a Daniell cell (Zn | Zn2+ || Cu2+ | Cu) if deionised water is added to the zinc sulfate half-cell beaker?

Show Answer & Explanation

Correct Answer: B

Lowering [Zn2+] shifts the equilibrium Zn2+ + 2e- ⇔ Zn to the left, releasing more electrons and making the zinc electrode potential more negative. This widens the gap between the two electrode potentials, increasing overall cell EMF.