The batteries in your remote and the engine in your car are only a couple of examples of how chemical reactions create power through the flow of electrons. The cell potential is the way in which we can measure how much voltage exists between the two half cells of a battery. We will explain how this is done and what components allow us to find the voltage that exists in an electrochemical cell.
The cell potential, \(E_{cell}\), is the measure of the potential difference between two half cells in an electrochemical cell. The potential difference is caused by the ability of electrons to flow from one half cell to the other. Electrons are able to move between electrodes because the chemical reaction is a redox reaction. A redox reaction occurs when a certain substance is oxidized, while another is reduced. During oxidation, the substance loses one or more electrons, and thus becomes positively charged. Conversely, during reduction, the substance gains electrons and becomes negatively charged. This relates to the measurement of the cell potential because the difference between the potential for the reducing agent to become oxidized and the oxidizing agent to become reduced will determine the cell potential. The cell potential (Ecell) is measured in voltage (V), which allows us to give a certain value to the cell potential.
An electrochemical cell is comprised of two half cells. In one half cell, the oxidation of a metal electrode occurs, and in the other half cell, the reduction of metal ions in solution occurs. The half cell essentially consists of a metal electrode of a certain metal submerged in an aqueous solution of the same metal ions. The electrode is connected to the other half cell, which contains an electrode of some metal submerged in an aqueous solution of subsequent metal ions. The first half cell, in this case, will be marked as the anode. In this half cell, the metal in atoms in the electrode become oxidized and join the other metal ions in the aqueous solution. An example of this would be a copper electrode, in which the Cu atoms in the electrode loses two electrons and becomes Cu2+ .
The Cu2+ ions would then join the aqueous solution that already has a certain molarity of Cu2+ ions. The electrons lost by the Cu atoms in the electrode are then transferred to the second half cell, which will be the cathode. In this example, we will assume that the second half cell consists of a silver electrode in an aqueous solution of silver ions. As the electrons are passed to the Ag electrode, the Ag+ ions in solution will become reduced and become an Ag atom on the Ag electrode. In order to balance the charge on both sides of the cell, the half cells are connected by a salt bridge. As the anode half cell becomes overwhelmed with Cu2+ ions, the negative anion of the salt will enter the solution and stabilized the charge. Similarly, in the cathode half cell, as the solution becomes more negatively charged, cations from the salt bridge will stabilize the charge.
For electrons to be transferred from the anode to the cathode, there must be some sort of energy potential that makes this phenomenon favorable. The potential energy that drives the redox reactions involved in electrochemical cells is the potential for the anode to become oxidized and the potential for the cathode to become reduced. The electrons involved in these cells will fall from the anode, which has a higher potential to become oxidized to the cathode, which has a lower potential to become oxidized. This is analogous to a rock falling from a cliff in which the rock will fall from a higher potential energy to a lower potential energy.
The difference between the anode's potential to become reduced and the cathode's potential to become reduced is the cell potential.
\[E^o_{Cell}= E^o_{Red,Cathode} - E^o_{Red,Anode} \nonumber \]
Note:
Here is the list of the all the components:
All of these components create the Electrochemical Cell.
The image above is an electrochemical cell. The voltmeter at the very top in the gold color is what measures the cell voltage, or the amount of energy being produced by the electrodes. This reading from the voltmeter is called the voltage of the electrochemical cell. This can also be called the potential difference between the half cells, Ecell. Volts are the amount of energy for each electrical charge; 1V=1J/C: V= voltage, J=joules, C=coulomb. The voltage is basically what propels the electrons to move. If there is a high voltage, that means there is high movement of electrons. The voltmeter reads the transfer of electrons from the anode to the cathode in Joules per Coulomb.
The image above is called the cell diagram. The cell diagram is a representation of the overall reaction in the electrochemical cell. The chemicals involved are what are actually reacting during the reduction and oxidation reactions. (The spectator ions are left out). In the cell diagram, the anode half cell is always written on the left side of the diagram, and in the cathode half cell is always written on the right side of the diagram. Both the anode and cathode are seperated by two vertical lines (ll) as seen in the blue cloud above. The electrodes (yellow circles) of both the anode and cathode solutions are seperated by a single vertical line (l). When there are more chemicals involved in the aqueous solution, they are added to the diagram by adding a comma and then the chemical. For example, in the image above, if copper wasn't being oxidized alone, and another chemical like K was involved, you would denote it as (Cu, K) in the diagram. The cell diagram makes it easier to see what is being oxidized and what is being reduced. These are the reactions that create the cell potential.
The standard cell potential (\(E^o_{cell}\)) is the difference of the two electrodes, which forms the voltage of that cell. To find the difference of the two half cells, the following equation is used:
\[E^o_{Cell}= E^o_{Red,Cathode} - E^o_{Red,Anode} \tag{1a} \]
with
The units of the potentials are typically measured in volts (V). Note that this equation can also be written as a sum rather than a difference
\[E^o_{Cell}= E^o_{Red,Cathode} + E^o_{Ox,Anode} \tag{1b} \]
where we have switched our strategy from taking the difference between two reduction potentials (which are traditionally what one finds in reference tables) to taking the sum of the oxidation potential and the reduction potential (which are the reactions that actually occur). Since E^o_{Red}=-E^o_{Ox}, the two approaches are equivalent.
The example will be using the picture of the Copper and Silver cell diagram. The oxidation half cell of the redox equation is:
Cu(s) → Cu2+(aq) + 2e- EoOx= -0.340 V
where we have negated the reduction potential EoRed= 0.340 V, which is the quantity we found from a list of standard reduction potentials, to find the oxidation potential EoOx. The reduction half cell is:
( Ag+ + e- → Ag(s) ) x2 EoRed= 0.800 V
where we have multiplied the reduction chemical equation by two in order to balance the electron count but we have not doubled EoRed since Eo values are given in units of voltage. Voltage is energy per charge, not energy per reaction, so it does not need to account for the number of reactions required to produce or consume the quantity of charge you are using to balance the equation. The chemical equations can be summed to find:
Cu(s) + 2Ag+ + 2e- → Cu2+(aq) + 2Ag(s) + 2e-
and simplified to find the overall reaction:
Cu(s) + 2Ag+ → Cu2+(aq) + 2Ag(s)
where the potentials of the half-cell reactions can be summed
EoCell= EoRed,Cathode+EoOx,Anode
EoCell = 0.800 V + (-0.340 V)
EoCell = 0.460V
to find that the standard cell potential of this cell is 0.460 V. We are done.
Note that since E^o_{Red}=-E^o_{Ox} we could have accomplished the same thing by taking the difference of the reduction potentials, where the absent or doubled negation accounts for the fact that the reverse of the reduction reaction is what actually occurs.
EoCell= EoRed,Cathode-EoRed,Anode
EoCell = 0.800V - 0.340V
EoCell = 0.460V
The table below is a list of important standard electrode potentials in the reduction state. To determine oxidation electrodes, the reduction equation can simply be flipped and its potential changed from positive to negative (and vice versa). When using the half cells below, instead of changing the potential the equation below can be used without changing any of the potentials from positive to negative (and vice versa):
EoCell= EoRed,Cathode - EoRed,Anode
Table: Reduction Half-Reaction Eo, V Acidic Solution F2(g) + 2e- → 2 F-(aq) +2.866 O3(g) + 2H+(aq) + 2e- → O2(g) + H2O(l) +2.075 S2O82-(aq) + 2e- → 2SO42-(aq) +2.01 H2O2(aq) + 2H+(aq) +2e- → 2H2O(l) +1.763 MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l) +1.51 PbO2(s) + 4H+(aq) + 2e- → Pb2+(aq) + 4H2O(l) +1.455 Cl2(g) + 2e- → 2Cl-(aq) +1.358 Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l) +1.33 MnO2(s) + 4H+(aq) +2e- -> Mn2+(aq) + 2H2O(l) +1.23 O2(g) + 4H+(aq) + 4e- → 2H2O(l) +1.229 2IO3-(aq) + 12H+(aq) + 10e- → I2(s) + 6H2O(l) +1.20 Br2(l) + 2e- → 2Br-(aq) +1.065 NO3-(aq) + 4H+(aq) + 3e- → NO(g) + 2 H2O(l) +0.956 Ag+(aq) + e- → Ag(s) +0.800 Fe3+(aq) + e- → Fe2+(aq) +0.771 O2(g) + 2H+(ag) + 2e- → H2O2(aq) +0.695 I2(s) + 2e- → 2I-(aq) +0.535 Cu2+(aq) + 2e- → Cu(s) +0.340 SO42-(aq) + 4H+(aq) + 2e- → 2H2O(l) + SO2(g) +0.17 Sn4+(aq) + 2e- → Sn2+(aq) +0.154 S(s) + 2H+(aq) + 2e- → H2S(g) +0.142H+(aq) + 2e-
→H2(g)
0
Pb2+(aq) + 2e- → Pb -0.125 Sn2+(aq) + 2e- → Sn(s) -0.137 Fe2+(aq) + 2e- → Fe(s) -0.440 Zn2+ + 2e- → Zn(s) -0.763 Al3+(aq) + 3e- → Al(s) -1.676 Mg2+(aq) + 2e- → Mg(s) -2.356 Na+(aq) + e- → Na(s) -2.713 Ca2+(aq) + 2e- → Ca(s) -2.84 K+(aq) + e- → K(s) -2.924 Li+(aq) + e- → Li(s) -3.040 Basic Solution O3(aq) + H2O(l) + 2e- → O2(g) + 2OH-(aq) +1.246 OCl-(aq) + H2O(l) + 2e- → Cl-(aq) + 2OH-(aq) +0.890 O2(g) + 2H2O(l) +4e- → 4OH-(aq) +0.401 2H2O(l) + + 2e- → H2(aq) + 2OH-(aq) -0.0828Eocell= 2.71V= +0.401V - Eo{Al(OH)4]-(aq)/Al(s)}
Eo{[Al(OH)4]-(aq)/Al(s)} = 0.401V - 2.71V = -2.31V
Confirm this on the table of standard reduction potentials
What is Galvanic Cell?
An electrochemical cell that converts the chemical energy of spontaneous redox reactions into electrical energy is known as a galvanic cell or a voltaic cell.
Galvanic cell Voltaic cell is an electrochemical cell that makes use of chemical reactions to generate electrical energy.
Let us understand how a voltaic or galvanic cell is created.
In oxidation-reduction reactions, electrons are moved from one species to another species. Energy is released if the reaction occurs spontaneously. Therefore, the released energy is used to do useful work. To tackle this energy, it is required to split the reaction into two separate half-reactions viz. oxidation and reduction. With the help of two different containers and wire, the reactions are put into them to drive the electrons from one end to the other end. This creates a voltaic cell.
Principle of
Galvanic (
Voltaic) Cell
Electric work done by a galvanic cell is mainly due to the Gibbs energy of spontaneous redox reaction in the voltaic cell. It generally consists of two half cells and a salt bridge. Each half cell further consists of a metallic electrode dipped into an electrolyte. These two half-cells are connected to a voltmeter and a switch externally with the help of metallic wires. In some cases, when both the electrodes are dipped in the same electrolyte, a salt bridge is not required.
Parts of
Galvanic
Cell
Anode – Oxidation occurs at this electrode.
Cathode – Reduction occurs at this electrode.
Salt bridge – Contains electrolytes which are required to complete the circuit in a galvanic cell.
Half-cells – reduction and oxidation reactions are separa
ted into compartments.
External circuit – Conducts the flow of electrons between electrodes
Load – A part of the circuit utilizes the electron to flow to perform its function.
Working of
Galvanic
Cell
Example of
Galvanic
Cell
Electrochemical or galvanic cells were introduced as a tool for studying the thermodynamic properties of fused salts more than a century ago. Daniel’s cell is an example of a galvanic cell that converts chemical energy into electrical energy. In Daniel’s cell, copper ions are reduced at the cathode while zinc is oxidized at the anode.
Reactions of Daniel cell at cathode and anode are:
At cathode: Cu 2+ + 2e– → Cu
At anode: Zn → Zn2+ + 2e–
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Frequently Asked Questions – FAQs
Q1
The electrochemical cell type is a galvanic cell. It is used to supply electrical current through a redox reaction to the transfer of electrons. A galvanic cell is an example of how to use simple reactions between a few elements to harness energy.
Q2
The basic cell or voltaic cell is made up of two electrodes, one of copper and the other of zinc dipped in a glass vessel solution of dilute sulfuric acid. The current flows from copper to zinc outside the cell and from zinc to copper inside the cell when the two electrodes are connected externally with a piece of wire.
Q3
The anode is the electrode where oxidation (loss of electrons) occurs; it is the negative electrode in a galvanic cell since electrons are left on the electrode when oxidation occurs. Therefore, the cathode is a neutral electrode; because there are reduced positive ions of metal atoms.
Q4
Sometimes known as a voltaic cell or Daniell cell is a galvanic cell. One example of a galvanic cell is the common household battery. The electrons flow from one chemical reaction to another occurs through an external circuit that results in the current.
Q5
A basic voltaic cell is created by immersing in a diluted solution of sulfuric acid a zinc plate and a copper plate. A voltaic cell is an electrochemical cell that produces electrical energy using a chemical reaction. The reactions to oxidation and reduction are divided into half-cell compartments.
Q6
A galvanic cell is an electrochemical cell that converts the chemical energy of spontaneous redox reactions into electrical energy. It is also known as a voltaic cell.
Q7
A galvanic cell needs a salt bridge to maintain the neutrality of the solution and allow the free flow of ions from one cell to another cell.
Q8
Oxidation occurs at the anode of a galvanic cell.
Q9
If the salt bridge dries out, the anode half will become positively charged, while the cathode half will become negatively charged. Due to this charge imbalance, the cell will get dead.
Q10
Nernst equation hypothesised the inverse relation of the voltage of the galvanic cell and the temperature, i.e. the voltage of the galvanic cell decreases with the increasing temperature.
For detailed discussions on different cells like a galvanic cell or voltaic cell, or about the Gibbs energy change spontaneity of a process visit BYJU’S.
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