Change of Potential Difference in Voltaic Cells Lab Report

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The Change of potential difference in Voltaic Cells

  1. Introduction

Oxidation-Reduction (redox) reaction is a group of reactions that are linked to the transfer of electrons between species. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Each reaction by itself is called a half-reaction because there must be two halves of reactions to form a whole reaction. When a redox reaction takes place, electrons are transferred from one species to the other.  If the reaction is spontaneous, energy is released.

A Voltaic Cell (Galvanic Cell) is an electrochemical cell that uses spontaneous redox reactions to generate electricity. It consists of two separate half-cells. A half-cell is composed of an electrode within a solution containing Mn+ ions in which M is the metal and n is the number of charges of the metal. The two half cells are linked together by a wire running from one electrode to the other with a voltmeter to measure the potential difference between the two electrodes. A salt bridge also connects to the half cells to keep solutions neutral and allow free flow of electrons. In the absence of a salt bridge, electrons won’t transfer and the voltmeter won’t measure any voltage. When an electrode is oxidized in a solution, it is called an anode and when an electrode is reduced in a solution it is called a cathode. The electrons will flow from the more negative half reaction to the more positive half reaction (i.e. from anode to cathode).  The readings from the voltmeter present the reaction's cell voltage (potential difference) between its two half-cells.

Standard Cell Potential: Eocell = Eo (cathode) - Eo (anode)

The Eo values of metals are tabulated with all solutes at 1 M and all gases at 1 atm. These values are called standard reduction potentials. Each half-reaction has a different reduction potential; the difference of two reduction potentials gives the voltage of the cell.

Research Question

How does the use of different concentrations of copper sulfate solution (half-cell solution) affect the potential difference in a Voltaic Cell?


A voltaic cell with a higher concentration of copper sulfate solution will increase the potential difference between the two half-cells.


Independent Variable

The use of different concentrations for Copper Sulfate solution; 0.2 M, 0.4 M, 0.6 M, 0.8 M, 1.0 M

Dependent Variable

The change in potential difference (voltage) between the two half-cells of a voltaic cell.

Controlled Variables

  • Salt Bridge (KCl), 1molar (M)

Different kinds of salt bridge solution can have an effect on the cell’s voltage or the ion movement.

  • Salt bridge length
  • Voltmeter

Using different kinds of Voltmeters can represent different readings.

  • The size of metal electrodes
  • The two metal electrodes (copper (Cu), and Zinc (Zn))

Different metal electrodes have an effect on the results of the cell’s voltage.

  • Zinc sulphate concentration, 1M
  • The volume of copper sulphate and zinc sulphate.
  1. Methods

Materials & Apparatus

  • Two 100 mL Glass Beakers
  • Filter papers
  • Glass rod
  • Two disposable gloves
  • Two Insulated connecting wires
  • Graduated cylinder (50 ml) ±0.5
  • Different concentrations of Copper (II) Sulfate solution (CuSO4): 0.2 M, 0.4 M, 0.6 M, 0.8 M and 1.0 M.
  • 1M of  Zinc (II) Sulfate solution (ZnSO4)
  • Volumetric flasks (50ml, 100ml, 250ml)
  • 1M of potassium chloride (KCl) solution ( i.e. for Salt bridge)
  • Voltmeter
  • Two metal electrodes: Zinc and copper (with exact dimensions).
  • Funnel
  • Distilled H2O


The data collected throughout the experiment will be displayed in a data table. The experiment will be repeated twice, so the data table will represent results for both trials and the average. The data tables include the copper sulfate concentrations dm-3mol (M) for each trial of the experiment, the potential difference between both half-cells in each copper sulfate concentration ±0.05 V. The data collected will be displayed in a scatter graph to represent the different concentrations of copper sulfate vs. the different voltages measured for each trial.

Part A

 Preparation of 1.0 M Copper Sulfate pentahydrate

 1. Find the mass of copper sulfate that is needed to prepare 50 mL (0.050 L) of 1.0 M solution using a 50mL volumetric flask. The molar mass of copper sulfate pentahydrate is 249.71g/mol.

 2. From the molar mass of copper sulfate pentahydrate, CuSO4.5H2O and the number of moles from Step 1, find the mass in grams of solute needed. Record your answer.

Mass = number of moles 1 molar mass

Mass of copper sulfate pentahydrate = 1  249.71 = 249.71 g

This mass is required to prepare 1L of 1M copper sulfate solution.

For 50ml, divide it by 20 = 12.485g of CuSO4.5H2O

3. Weigh 12.485g of CuSO4.5H2O using a 0.01 digital scale, and carefully put the blue crystals in the 50 mL volumetric flask using the funnel.

4. Add about half the volume (25ml) of distilled water needed and swirl the flask. When most of the solid has dissolved add the rest of the water stopping below the mark on the flask. To add the remaining water use the water wash bottle. In this way, the 1M CuSO4 solution is ready to use.

Throughout this experiment, 5 different concentrations of Copper Sulfate pentahydrate should be prepared to measure the change in voltage. The different concentrations that we are going to use are: 0.2 M, 0.4 M, 0.6 M, 0.8 M and 1.0 M.  Repeat these four steps for preparing the other concentrations of copper sulfate pentahydrate throughout the experiment. The different masses of CuSO4.5H2O for different concentrations can be calculated in the same way, using molar mass 249.71g/mol for CuSO4.5H2O.

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It is easier to prepare a stock solution of 1dm-3mol (1M) copper sulfate solution and then use the dilution method to prepare the different concentrations (see sample calculations)

Preparation of 1.0M Zinc Sulfate hepta-hydrate solution:

The same calculation can be done to find the mass of zinc sulfate which is needed to prepare 1dm-3mol (1M) ZnSO4 solution.

1. Find the number of moles needed to make 50 mL (0.050 L) of 1.0 M solution using a 50mL volumetric flask. Record your answer.  Molar mass of zinc sulfate hepta-hydrate is 287.53g/mol

 2. From the molar mass of zinc sulfate ...

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