Order of Half Cells in The Electrochemical Series

Order of half cells in the electrochemical series
Title: How to determine the ranking electrochemical series by using the chemical reaction of
half cells.
Aim: To design and conduct an experiment to determine the order of four half-equations in
the electrochemical series
Introduction:
An electrochemical cell is based on an oxidation-reduction, which is a redox reaction and
consists of two half-cells: an anode half-cell and a cathode half-cell. Oxidation occurs at the
anode; reduction occurs at the cathode. An electrochemical cell can produce an electric
current, which is driven by an electrical potential difference between the two half-cells.
Just which electrode is negative, and which is positive in a particular electrochemical cell
depends upon the relative ability of the various chemical species in the half-cells to donate
and accept electrons. The half-cell containing the stronger reductant will undergo an
oxidation reaction; the electrode in this half-cell will be negative. The other half-cell will
contain the stronger oxidant and undergo a reduction reaction. In this experiment the
experimenter will use a volumeter to measure and hence compare the electrical potential
differences of several electrochemical cells and construct an electrochemical series order.
Variables: Independent variable: the different chemical species (conjugate pairs) in the
unknown half-cells
Dependent variable: the voltage
Controlled variables: same temperature, same electrolyte concentration, same volumeter.
Material: Safety measures:
50 mL 0.1 M copper (II) nitrate solution
50 mL 0.1 M zinc nitrate solution
50 mL of an equal mixture of 0.5 M iodine and 0.1 M potassium iodide solution
50 mL of an equal mixture of 0.1 M iron (III) nitrate and 0.1 M iron (II) sulfate solution
50 mL 0.1 M potassium nitrate solution
Copper electrode
Zinc electrode
2 carbon rods
5 x 100 mL beakers
Voltmeter
2 wire leads with alligator clips
6 filter papers, 12.5 cm Whatman No. 1
Safety glasses
Lab coat
Gloves
Eye protection
The overall cell potential can be regarded as the sum of the two half-cell potentials:
E_cell = E_cathode – E_anode
Methodology:
Construct the following half-cells:
〖𝐶𝑢〗^(2+)/〖𝐶𝑢〗^ – 100 mL beaker containing a copper electrode and 50 mL copper (II)
nitrate solution
〖𝐹𝑒〗^(3+)/〖𝐹𝑒〗^(2+) - 100 mL beaker containing a carbon rod and 50 mL iron (III)
nitrate/iron (II) sulfate solution
𝐼_2/𝐼^− - 100 mL beaker containing a carbon rod and 50 mL iodine/potassium iodide
solution
〖𝑍𝑛〗^(2+)/〖𝑍𝑛〗^ - 100 mL beaker containing a zinc electrode and 50 mL zinc nitrate
solution
Make a salt bridge by soaking a strip of filter paper in a beaker of potassium nitrate solution.
Join the 〖𝐶𝑢〗^(2+)/〖𝐶𝑢〗^ and 〖𝑍𝑛〗^(2+)/〖𝑍𝑛〗^ half-cells using the salt bridge.
Attach two wires to the voltmeter and clip the other end of one wire to the copper
electrode. Momentarily touch the loose end of the second wire to the zinc electrode. If the
needle of the voltmeter is deflected onto the scale, clip this loose end of the wire to the zinc
electrode and proceed with the next step. If the needle of the voltmeter is deflected below
zero, swap the wires at the terminals of the voltmeter before continuing.
Record the voltage and identify the positive and negative electrodes. (The positive electrode
is connected to the positive terminal of the voltmeter).
Disconnect the leads. Remove and discard the salt bridge.
Repeat steps 2 – 4 for each of the other five combinations of half-cells
Result:
Electrochemical cell Negative

potential (V) Cu|Cu 2+
Zn|Zn 2+
I^-|I2 Fe|Fe2+
Cu|Cu2+ 0.951 0.282 0.572
Positive lead
Zn|Zn2+ 0.951 1.221 1.521

I^-|I2 0.282 1.221 0.284
Fe|Fe2+ 0.521 1.521 0.284
The order of the electrochemical series produced:

_____Zn_____ > ____Fe______ > ____Cu____>____I^-____
Salvanic Diagrams:
Discussion:
In the illustrations for this experiment, the four half cell reactions are described as reduction
reactions in the diagram given above. Whereas Zinc and Iodine are discovered to be the
greatest oxidant and reductant, respectively. The technique employed to obtain these
results involved evaluating overall cell potentials, which are the total of two half-cells. In
descending order, the list is as follows: Zn_____ > ____Fe_____ > ____Cu > I-____. These
commands are checked with the electrochemical series from the data book to inspect for
any possible anomalies. Considering the two-ranking series, it
is shown that the two are practical for considering in a two-ranking series. Therefore, there
is no significant error that could have affected the results of this study. In a scenario of an
alternative half-cell given: Pb^2+|Pb, if Pb is a stronger reductant than copper, but Pb^2+ is
a stronger oxidant than Zn^2+, where in the electrochemical series would it be placed? The
answer would be between Fe and Cu. It is believed that the main function of salt bridges in
an electrochemical cell is to balance charges when electrons move from one half of the cell
to the other half of the cell. During this process, the salt bridge uses its electrolyte solution
which further helps in balancing charges in both the half cells. As the ions pass through the
salt bridge, it is possible that a few still remain attached.. If this salt bridge is used in another
electrochemical cell, then the attached ions will interrupt the reaction. Therefore, a new salt
bridge is necessary.
Conclusion:
It may be determined that galvanic cells are capable of helping establish the sequence of
four half-equations in the electrochemical series. As an outcome, the hypothesis has been
confirmed.

Order of Half Cells in The Electrochemical Series

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Order of half cells in the electrochemical series

Electrochemical cell Negative

Zn|Zn2+ 0.951 1.221 1.521

The order of the electrochemical series produced:

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