Electrochemistry
Writing Half Equations and Conventional Representation of Cells.
Electrochemical cells can be though of as performing a reaction in two separate phases with a movement of electrons through an external circuit which can be used as a power source.
A Daniel cell is an electrochemical cell containing zinc and copper, a more familiar reaction would be the displacement reaction between zinc and copper sulfate.
Electrochemistry can be split in to the following sections;
· Writing Half-Equations and conventional representation of half-cells.
· The Standard Hydrogen Electrode and measuring EMF
· Non-standard conditions for cells
· Redox reactions
· Applications of electrochemistry.
By removing the spectator ion we have the ionic equation
This ionic reaction shows the transfer of electrons from the zinc to the copper and can be further split in to the following half equations.
In a conventional reaction, as we learnt in kinetics with collision theory, the reactants must collide together in order for a reaction to occur. In this case an electron transfer from the zinc to the copper.
In a Daniel cell, and other commercial batteries, this electron transfer from zinc to copper occurs through a wire, this movement of electrons is a current and can be used to power electrical devices.
Writing half equation for each half cell is initially the same approach as writing half equations from the Redox section. Step 2 and 4 are omitted if there is no oxygens bound to the species oxidised/reduced.
1. Balance the species being oxidised/reduced
2. Balance the oxygen by adding waters
3. Balance the hydrogen by adding protons
4. Balance the electrons by using the oxidation state of species oxidised/reduced
The equation is then written with the addition of the electrons on the left hand side.
In the half equations for the Daniel cell can be written as electrode potentials, the above half equation for the Zn will need to be reversed so the electrons are on the left hand side, the reactants.
A convention for writing cells has been produced by IUPAC just like naming conventions for organic molecules. As water is a solvent it is omitted in the conventional cell.
| represent a phase boundary
|| represents a salt bridge This is used to “complete the circuit”
The most oxidised species are written next to the salt bridge and then written in increasing oxidation state.
A Platinum electrode must be included if there is no solid present in the half equation which are written furthest from the salt bridge. Platinum is used as it is a very unreactive metal so will not interfere with the reaction. Electrodes are needed to provide a surface for the reaction.
State symbols need to be included.
For example the half cell for the equation
Zn2+(aq) + 2e- —> Zn(s)
The most oxidised species is Zn2+ so must be drawn next to the salt bridge.
Zn2+(aq)||
The Zn(s) is then written further from the salt bridge and as it is in a different phase must be separated from the Zn2+ with | indicating phase boundary.
Zn(s)|Zn2+(aq)||
There is a solid component of the cell, Zn(s), so no platinum electrode is needed.
A more complex example,
MnO4- + 8H+ + 5e- —> Mn2+ + 4H2O, the electrons are on the left so does not need to be switched.
The most oxidised is MnO4- so goes next to the salt bridge.
MnO4-(aq) ||
The next most oxidised species is Mn2+(aq), it is in the same phase as MnO4, so its going to be separated by a comma.
Mn2+(aq),MnO4-(aq) ||
We now add the final ion, the proton which is also in the same phase
H+(aq), Mn2+, MnO4-(aq) ||
There is no metal present to use as an electrode so we must use a platinum electrode which is added to the outside, it will always be in a different phase.
Pt(s) | H+(aq), Mn2+, MnO4-(aq) ||
The Standard Hydrogen Electrode and Measuring EMF
Each half cell will produce an electrode potential, this voltage is referred to as the electromotive force which is represented as E. If it is in standard conditions, which we will mostly be dealing with the symbol;
This symbol is electromotive force in standard conditions measured in Volts.
All potential differences must be measured between two points, for our electromotive force, E, we measure it against a Standard Hydrogen Electrode which has a E/V of 0.00 by definition.
The Standard Hydrogen Electrode is a H+/H2 electrode under standard contions, the standard conditions for a half cell are
· 298K
· Ions 1moldm-3
· Pressure of Gas 100kPa
The equation for the Standard Hydrogen Electrode is
There is no metal present for an electrode so we need to add a platinum electrode for a reaction surface, so the standard notation for the Standard hydrogen Electrode is,
Pt(s)|H2(g)|H+(aq)||
The protons, H+, need to have a concentration of 1mol/dm3, this could be achieved with a 1mol/dm3 solution of HCl or 0.5mol/dm3 solution of H2SO4. This is because H2SO4 is diprotic.
Using The Standard Hydrogen Electrode
To measure EMF a Standard hydrogen electrode is connected to a half cell using a voltmeter and a salt bridge. The salt bridge is a component, often paper, which is saturated with ions which can carry the charged ions and therefore completes the circuit. The ions used are often potassium nitrate as they do not interact with the other species present. Voltmeters ideally have an infinite resistance so the current around the completed cell is zero, if there was a current the concentration of ions would change and no longer be standard.
Measuring EMF
The standard electrode potentials can then be listed as an electrochemical series.
The most positive E/V is written at the top, this shows the species which is most likely to accept an electron, the most powerful oxidising agent. In the list above F2 is the most powerful oxidising agent. The most negative at the bottom identifies the most powerful reducing agent, on this list it is Li. Note that we must select the appropriate side of the equation, a reducing agent provide electrons so the electron must be in the products so the equation is reversed.
The most negative Eo/V indicates where the electrons are going to flow from.
Combining Half cells
For example we are going to calculate the EMF of a Daniel cell
To combined half cells and calculate the total EMF we need to draw the standard half cells
Cu(s)|Cu2+(aq)||
Zn(s)|Zn2+(aq)||
To combine these we need to put on of the cells on the left hand side of the salt bridge, we put the half cells in numerical order. Most negative/least positive on the left hand side and least negative/most positive on the right hand side. This is a standard cell notation, they can be non-standard where they put them in the opposite position, calculating EMF for standard and non-standard cell notations are the same.
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
To calculate the EMF of a cell we do the calculation
(EMF right hand cell) – (EMF left hand cell) = Cell EMF
0.34-(-0.76) = 1.1V
E/V Values for non standard cells.
E values effectively indicate how willing a species is to be reduced, another way to think of it is the most negative electrode will produce the electrons. In the Daniel cell the Zn/Zn2+ electrode has a more negative E/V value so the electrons will come from that half cell. All of the half cells are effectively in equilibrium so the production of electrons is also in equilibrium and anything which would affect an equilibrium position will affect the E/V value.
If the above half cell is at 300K instead of the standard conditions of 298K there will be an affect on the E/V of the cell. If the temperature increases then Chatelier’s principle tells use that the system will try to resist the change and the endothermic reaction will be performed at an increased rate. In the case of the above equilibrium the backwards reaction is endothermic so the equilibrium position will shift to the left meaning an increase in the production of electrons. With an increase in the amount of electrons the E/V value will be more negative.
If a concentration of Zn2+ is increased we can use Chatelier’s principle again which will mean the equilibrium will shift to the right hand side and the amount of electrons will reduce. This will make the E/V of the cell be less negative/more positive.
If the platinum electrode is enlarged there is a greater surface area for the reaction to take place, so both reactions will increase so the E/V value will remain the same as the equilibrium position remains constant. Think of it similar to adding a catalyst to an equilibrium.
Answering Non-Standard Cell questions
a) For the below cell in standard conditions calculate the overall E/V
Pb(s)|Pb2+(aq)||Cu2+(aq)|Cu(s)
Right Hand Side – Left Hand Side = Cell EMF
0.34 – (-0.13) = 0.47V
b) What effect if any will increasing the concentration of Pb2+ ions have on the cells E/V
To answer this question we need to work out if the Pb half cell will change in E/V, the Cu half cell will not be affected.
The equilibrium of the Pb cell is Pb2+(aq) + 2e- —> Pb(s)
If we increase the concentration of Pb2+ ions the equilibrium position will shift right and decrease the amount of electrons so the half cell will become less negative. We could imagine it is -0.12
0.34 – (-0.12) = 0.46V, The E/V has reduced.
Redox reactions
Some redox reactions we have been predicting the reaction of many years before we have heard of redox, for example a reaction between chloring and potassium iodide. This is a very simple redox we will use to show the usefulness of the electrochemical series.
The E/V values tell us that the electrons will come from the iodine half cell, the most negative/least positive and go to the chlorine half cell as it is the most positive. The oxidizing agent will be Cl2 and the reducing agent will be I-.
The iodine half cell equation needs to be switched as it is providing the electrons.
The above half equations can be combined in to the ionic equation.
The electrochemical series can be used to determine if a species can oxidise another, if a species is higher on the electrochemical series it can oxidise something below it.
Answering Redox Questions
From the above electrochemical series identify a reducing agent that will reduce Vanadium (V) Oxide to Vanadium (IV)+ but no further.
We first need to identify the equation which is the Vanadium (V) Oxide being reduced to Vanadium (IV) Oxide, this is the 6th equation with a E/V value of +1.00, to reduce the species we need a half cell which has a lower E/V so it can provide the additional electrons. Equations 1-5 have a lower E/V so could contain the reducing agents, but we now need to find the equation which produces Vanadium (III). The 4th equation shows the production of Vanadium (III) with and E/V of 0.34, so anything with an E/V value of 0.34 or less will reduce Vanadium (V) oxide to Vanadium (III). There is only one hale cell with an E/V value between 1.00 and 0.34.
Fe3+ + e- —> Fe2+ +0.77
We now need to identify the reducing agent, the one providing the electrons, which is Fe2+. So the reducing agent which can reduce Vanadium (V) Oxide to Vanadium (IV) Oxide and no further is Fe2+.
Applications of Electrochemistry
Electrochemical cells can be used as commercial sources of energy in the form of cells or batteries, A common portable battery is the Lithium Ion Cell
Some cells can be recharged, this means the overall equation is reversable. By providing a potential difference in the opposite direction to its discharge the equation will go backwards and “recharge” the cell by increasing the concentration of reactants.
Hydrogen fuel cells are cells which generate a electricity using the oxidation of hydrogen gas using oxygen from the air.
The hydrogen fuel cell is not rechargeable, it is not plugged in to recharge, it is simply refuel with more hydrogen. The waste product of the hydrogen fuel cell is water so could potentially be non-harmful to the environment depending on the source of hydrogen. It is often made from crude oil, but could be made in alternative methods like electrolysis of water. Hydrogen is very difficult to store as it is a gas and it is explosive, these problems currently reduce the uptake of hydrogen fuel cells.
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