Chemistry Report: Measuring Standard Electrode Potential of Iron

This report focuses on measuring the standard electrode potential of Iron III ions using a standard hydrogen electrode. It covers the underlying theory of electrode potentials, including redox reactions, half-reactions, and the Nernst equation. The report details the experimental setup and procedure for measuring the electrode potential, explaining the role of the standard hydrogen electrode as a reference. It includes calculations of cell potential, Gibbs free energy, and equilibrium constants for reactions involving Iron III and other half-cells. The report also discusses the solubility of salts in terms of Gibbs free energy and temperature dependence. The student used data to determine the solubility of various salts at room temperature and calculated the temperature at which solubility becomes unfeasible. The document is available on Desklib, a platform offering a range of study tools and solved assignments for students.

Assignment 5: Redox 1
University Affiliation
Assignment 5: Redox
By
Name of Student
Date
Name of Lecturer

Assignment 5: Redox 2
Question 1
Report On how to measure a standard Electrode potential of Iron III ions
Introduction
Standard electrode potential is basically a measurement of equilibrium potential
Handbook of electrochemistry, 2014). When certain conditions such as temperature,
concentration are changed, this equilibrium often shifts. Hence, standard condition must be used.
The standard conditions usually uses are:
Pressure=101.3 kPa(1 atm)
Temperature=298 k ∨(25 ℃)
Concentration=1 mol . dm−3
The Standard Hydrogen Electrode
This is the potential difference between the two electrodes, which contributes to the flow
of electron from the anode to cathode via galvanic cell, which forms an external circuit (Clugston
& Flemming, 2015).
The potential of the electrolyte and electrode can be measure though the process is quite
involving and the value obtained depends on pressure, temperature and electrolyte solution
concentration. A method to eliminate all these variations involves using a standard reference
electrode to compare any electrode potential (Moody, 2014). Such comparison is performed is a
predetermined condition of same pressure, temperature and concentrations. It implies that such
values can help in calculating the potential difference between two electrodes under
investigation. In addition, it implies that you need not to construct a particular cell in order to
compare the electrode potential differences (Skoog, Holler & Crouch, 2016).

Assignment 5: Redox 3
Standard hydrogen electrode
The standard hydrogen electrode is basically a universal for finding any half-cell potentials
because it is a redox electrode.
Fig 1: a standard hydrogen electrode (Clark et al., 2013).
The standard hydrogen electrode comprises a platinum electrode in a solution having
H+¿¿ ions. For example, H2 S O4 (aq) with a concentration of 1 mod . dm−3 and the following
equation describes what occurs.
2 H+¿(aq)+2 e−¿⇌ H 2(g) ¿¿
Hydrogen gas is released in the electrodes
The hydrogen electrode is only used in the experiment only and only if it can be attached
to the electrode system under investigation. In this, our objective is to determine the electrode

Assignment 5: Redox 4
potential of Iron III, hence, the Iron III half cell will be connected to the hydrogen electrode to
determine the electrode potential of iron III (Skoog, Holler & Crouch, 2016).
By convention, all the tabularized values of standard electrode potential are listed for a
reduction reaction and not as oxidation, to allow for standard potentials comparison for various
substances. The standard potential cell potential Ecell
o is thus the difference cathode value and
anode value in the two half-cell reactions (Skoog, Holler & Crouch, 2016).
Ecell
o =Ecathode−Eanode
Half-reactions usually show the oxidation and reduction reactions occurring in the cells.
Adding the two half reactions gives the overall reaction. When a standard potential of one-half
reaction is known, it is possible to determine the value of standard potential of other half-
reactions. This is achieved by measuring the standard potential of the corresponding cell.
(Clugston & Flemming, 2015).
Fig 2
Objective
The objective of the experiment was to measure standard Electrode potential of Iron III irons

Assignment 5: Redox 5
Theory
Iron is silvery-white when its pure form. Iron is a metal. In ion form, it can exist in the
form of ion III ions and ion II ions. Iron is a magnetic material and cam dissolve small quantities
of carbon when molten thus yielding steel (Clugston & Flemming, 2015). Iron is commercially
refined from Fe2O3 or Fe3O4 also known as magnetite mixed with other substances at a very
high temperature of the blast furnace. The oxide are normally reduced to pure iron. Besides
hardening the iron through the addition of carbon and other metals to the molten iron, iron
forgings or iron castings can be treated by heat to exploit the different physical properties of
various phases of solid iron. Pure iron will readily reacts with moisture and oxygen in the
environment to produce rusts a process that is accompanied by corrosion (Clugston & Flemming,
2015).
A chemical reaction that involves electrons transfers from a given reactant to another
reactant is known as redox reactions or oxidation-reduction reactions. Two half-reactions usually
occur with one reactant donating an electron while the other gaining the electron. The reactant
gaining the electrons undergoes reduction while the one that donate the electrons undergoes
oxidation. A reduction process involves the reduction or decreases in oxidation number of a
substance. Chemical equations that represent half-reactions have to be both balanced in mass and
charge (Clugston & Flemming, 2015).
All half reaction can be expressed as reductions. A tendency measure for reduction to
take place is its reduction potential (E), which is measured in volts. A standard conditions of
Pressure=101.3 kPa(1 atm), Temperature=298 k ∨(25 ℃) and Concentration=1 mol . dm−3 is

Assignment 5: Redox 6
required for the half reaction. A half reduction half-reaction. Standard reduction potential is
therefore the tendency where a chemical species is often reduced and it unit is volts. A less
negative or more positive reduction potential indicated higher tendency of reduction taking place
(Clugston & Flemming, 2015).
A galvanic or voltaic cells is used to measure standard reduction potential. Under the set-
up, a redox reaction spontaneously occurs and an electric current is thus produced. For electron
transfers in a given redox reaction to produce a resultant electric current, electrons are usually
made to pass via an external electrical wire instead of flowing directly between the reducing and
oxidizing agents (Clugston & Flemming, 2015).
A galvanic cell of the two solution, one solution must contain the ions of the oxidation
half-reaction while their must contain the reduction of half-reaction ions. The two reactants are
put in separated cubicles termed as half-cells. An electrode is placed in for every a half cell in the
solution and connected externally to electrical wire. The anode is the end of one of the electrode
where oxidation happens while the cathode is one of the ends of the electrode in which reduction
occurs. (Clugston & Flemming, 2015). A salt bridge connects the two half-cells and it lets the
current out of the ions from a given half-cell to the next electrode to complete the electron
current circuit within the external wires. The circuit is complete when the two electrode are
connected to a load with electrical properties such as voltmeter, or a light bulb and
reduction/oxidation reaction takes place and the electron flows from the anode (negative) to
cathode (positive) thus producing electric current.
Apparatus
1. Two standard electrodes
2. Two beakers

Assignment 5: Redox 7
3. Voltmeter
4. Salt bridge
5. Fe3+ solution
6. H+ solution
7. Electrical wire
Procedure
To measure standard electron potential of iron III ions, the step up in figure 3 below was
arranged.
Fig 3

Assignment 5: Redox 8
Fig 3 above shows how the half-cell potential for Iron III ions can be measured during its
introversion. The theoretical value expected is +0.77. It means the electrons flows from negative
pole of the half-cell of H+¿¿ half-cell to positive pole of Fe+2 / Fe+3
Fe+3 ( aq )+ e−¿Fe+2 (aq)¿
The cell potential is given by:Ecell
o =Ecathode−Eanode and is a voltage measure that a battery can
provide. It is calculated mainly form the half-cell reduction potentials. The voltmeter placed
between the electrode will measure the Ecell
o . For Fe3+/Fe2+ the value is supposed to be +0.77
Question 2
Given that the measurement for the cell is: Eᶱ = +0.77V, describe the reactions (in terms of
species oxidised and reduced) taking place when the Fe3+/Fe2+ half-cell is connected to the
following ones:
MnO4
−¿ ( aq ) +8 H+¿ ( aq ) + 5 e−¿ → Mn+2 ( aq ) +4H 2O ( l) E o=+1.51 V ¿
¿¿
Answer : Mn O4
−¿ ¿ ion electrode potential is more positive than that of Fe+3 / Fe+2 hence
Mn O4
−¿ ¿ will be more readily reduced than Fe+3 / Fe+2 and Fe+3 / Fe+2 will be more readily
oxidized than Mn O4
−¿ ¿
Hence Mn O4
−¿ ¿ is reduced
Mn O4
−¿+8 H+ ¿+ 5e−¿→ Mn +2+4 H2O ¿
¿¿
while Fe2+ I is oxidized as indicated in the equation below:

Assignment 5: Redox 9
Fe+2 → Fe+3 +e−¿ ¿
Therefore
Mn O4
−¿ (aq )+8 H + ¿ ( aq )+ 5e−¿ →Mn +2 ( aq ) +4H 2O (l) ¿
¿¿
5 Fe+2 (aq)→5 Fe+3 (aq)+5 e−¿¿
---------------------------------------------------------------------------------------------------------
Mn O4 (aq)
−¿+8 H+ ¿ ( aq ) +5 Fe+2 ( aq ) →Mn+2 ( aq ) +5 Fe+3 (aq )+ 4 H2 O(aq )¿ ¿
Since Fe3+/Fe2+ is more positive than Cu+ it is reduced while Cu+ oxidized
Cu+ (aq) + e-  Cu(s) Eᶱ = +0.52V
2 Fe+3 (aq)+2 e−¿→ 2 Fe+ 2(aq)¿
Cu ( s ) → Cu+2(aq)+2 e−¿¿
----------------------------------------------------------------------------------------------
2 Fe+3 ( aq ) +Cu ( s ) → 2 Fe+2 ( aq ) +Cu+2 (aq)
ii) Write balanced equations for the reactions taking place in each cell when current is allowed to
flow.
Mn O4
−¿+8 H+ ¿+ 5Fe +2→ Mn+ 2
+ 5 Fe+3+ 4 H 2O (l)¿ ¿ (1)
2 Fe+3 ( aq ) +Cu ( s ) → 2 Fe+2 ( aq ) +Cu+2 (aq) (2)
iii) Calculate Ecell for both the cells formed.

Assignment 5: Redox 10
Ecell
o =Ecathode−Eanode
Ecell
o =+1.51 V −+0.77 V =+ 0.74
Ecell
o =+0.77 V −+0.52V =+0.25
iv) Calculate the value of the Gibbs free energy, G for each cell in jmol-1. Use the values for
Ecell and G to comment on the relative feasibility of these reactions. Explain why the free
energy is never fully obtained in practice from a working cell.
∆ G=−nF ECell
o (Schüring et al., 2015; Barrett, 2013).
For the first case:
Ecell
o =0.74
n=5
F=96458 C /mol e−¿¿
G=−5 mol e−¿× 96458 C
mol e−¿ ×0.74=−356894.6 j ¿¿
For the second case:
G=−2m ol e−¿×96458 C
mol e−¿× 0.25=−48229 j¿ ¿
since Eocell is positive in both cases and ∆G is negative, the reactions are spontaneous (Reger,
Goode & Ball, 2015; Masterton & Hurley, 2013; Cockell, 2015).
v) Hence calculate the equilibrium constant for each cell under these conditions. What do the
values suggest about the extent of each reaction?
Taking=298 k , RT is a constant then E ° cell=(0.025693V /n)
(i) K=e0.74 × 5
0.025693

Assignment 5: Redox 11
K=e143.98
(ii) Ii K=e0.25× 2
0.025693
k =e38.91
Question 3
Solubility of salts can be understood in terms of G using the relationship: G= H -TSsys.
i) Use the information in the table to determine which, if any, of the salts are soluble at
room temperature. [3.3]
Salt Hsol(kj mol-1) Ssys (j mol-1)
LiF +5 -37
MgSO4 -92 -213
NH4Cl +15 +167
CaSO4 -18 -145

Assignment 5: Redox 12
G= H -TSsys
∆ GLiF =5−−37=42
∆ GMgSO 4❑
=−92−−213=121
∆ GNH4 Cl=15∓167=−152
∆ GCaSO 4
=−18−−145=42
It implies that only NH4Cl is highly soluble in water at room temperature. Since Gibbs value
is negative, the process is endothermic and it is soluble (Atkins, 2014). The equilibrium
solubility thus depends of Gibbs value, negative values of Gibbs shows good solubility of salt
that is to provision of a high concentration of ± ion charge carriers and good conductance
(Conway, 2013; Weller et al., 2018).
II). For any salts above that are soluble at room temperature, use the equation to determine at
what temperature solubility just becomes unfeasible. Comment on the practical implications of
this given the liquid range of water.[3.4]
For ammonium Chloride
G=H −TS
−152=15−167 T
T =−167
167 =1 ℃
This implies that the salt is highly soluble in water and the process is endothermic where
ammonium chloride absorbs heat from water molecules hence the low temperature observed.