Voltaic Cells Galvanic cell
Saturday, March 23, 2019

# Galvanic cell

Galvanic cell with no cation flow

A galvanic cell or voltaic cell, named after Luigi Galvani or Alessandro Volta , respectively, is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within the cell. It generally consists of two different metals immersed in an electrolyte, or of individual half-cells with different metals and their ions in solution connected by a salt bridge or separated by a porous membrane.

Volta was the inventor of the voltaic pile , the first electrical battery .

In common usage, the word “battery” has come to include a single galvanic cell, but a battery properly consists of multiple cells. [1]

## Contents

• 1 History
• 2 Basic description
• 3 Energetics driving galvanic cell reactions
• 4 Half reactions and conventions
• 5 Cell voltage
• 6 Galvanic corrosion
• 7 Cell types
• 9 References

## History[ edit ]

In 1780, Luigi Galvani discovered that when two different metals (e.g., copper and zinc) are in contact and then both at the same time to two different parts of a muscle of a frog leg, to close the circuit, the frog’s leg contracts. [2]
He called this ” animal electricity “. The frog’s leg, as well as being a detector of electrical current, was also the electrolyte (to use the language of modern chemistry).

A year after Galvani published his work (1790), Alessandro Volta showed that the frog was not necessary, using instead a force-based detector and brine-soaked paper (as electrolyte). (Earlier Volta had established the law of capacitance C = Q/V with force-based detectors). In 1799 Volta invented the voltaic pile, which is a pile of galvanic cells each consisting of a metal disk, an electrolyte layer, and a disk of a different metal. He built it entirely out of non-biological material to challenge Galvani’s (and the later experimenter Leopoldo Nobili )’s animal electricity theory in favor of his own metal-metal contact electricity theory. [3] Carlo Matteucci in his turn constructed a battery entirely out of biological material in answer to Volta. [4] Volta’s contact electricity view characterized each electrode with a number that we would now call the work function of the electrode. This view ignored the chemical reactions at the electrode-electrolyte interfaces, which include H2 formation on the more noble metal in Volta’s pile.

Although Volta did not understand the operation of the battery or the galvanic cell, these discoveries paved the way for electrical batteries; Volta’s cell was named an IEEE Milestone in 1999. [5]

Some forty years later, Faraday (see Faraday’s laws of electrolysis ) showed that the galvanic cell — now often called a voltaic cell — was chemical in nature. Faraday introduced new terminology to the language of chemistry: electrode ( cathode and anode ), electrolyte , and ion ( cation and anion ). Thus Galvani incorrectly thought the source of electricity (or source of emf, or seat of emf) was in the animal, Volta incorrectly thought it was in the physical properties of the isolated electrodes, but Faraday correctly identified the source of emf as the chemical reactions at the two electrode-electrolyte interfaces. The authoritative work on the intellectual history of the voltaic cell remains that by Ostwald. [6]

It was suggested by Wilhelm König in 1940 that the object known as the Baghdad battery might represent galvanic cell technology from ancient Parthia . Replicas filled with citric acid or grape juice have been shown to produce a voltage. However, it is far from certain that this was its purpose—other scholars have pointed out that it is very similar to vessels known to have been used for storing parchment scrolls. [7]

## Basic description[ edit ]

Schematic of Zn-Cu galvanic cell

In its simplest form, a half-cell consists of a solid metal (called an electrode ) that is submerged in a solution; the solution contains cations (+) of the electrode metal and anions (−) to balance the charge of the cations. The full cell consists of two half-cells, usually connected by a semi-permeable membrane or by a salt bridge that prevents the ions of the more noble metal from plating out at the other electrode.

A specific example is the Daniell cell (see figure), with a zinc (Zn) half-cell containing a solution of ZnSO4 (zinc sulfate) and a copper (Cu) half-cell containing a solution of CuSO4 (copper sulfate). A salt bridge is used here to complete the electric circuit.

If an external electrical conductor connects the copper and zinc electrodes, zinc from the zinc electrode dissolves into the solution as Zn2+ ions (oxidation), releasing electrons that enter the external conductor. To compensate for the increased zinc ion concentration, via the salt bridge zinc ions leave and anions enter the zinc half-cell. In the copper half-cell, the copper ions plate onto the copper electrode (reduction), taking up electrons that leave the external conductor. Since the Cu2+ ions (cations) plate onto the copper electrode, the latter is called the cathode. Correspondingly the zinc electrode is the anode. The electrochemical reaction is:

Zn + Cu2+ → Zn2+ + Cu

In addition, electrons flow through the external conductor, which is the primary application of the galvanic cell.

As discussed under #Cell voltage , the emf of the cell is the difference of the half-cell potentials, a measure of the relative ease of dissolution of the two electrodes into the electrolyte. The emf depends on both the electrodes and on the electrolyte, an indication that the emf is chemical in nature.

## Energetics driving galvanic cell reactions[ edit ]

The electrochemical processes in a galvanic cell occur because reactants of high free energy (e.g. metallic Zn and hydrated Cu2+ in the Daniell cell) are converted to lower-energy products (metallic Cu and hydrated Zn2+ in this example). The difference in the lattice cohesive energies [8] of the electrode metals is sometimes the dominant energetic driver of the reaction, specifically in the Daniell cell. [9] Metallic Zn, Cd, Li, and Na, which are not stabilized by d-orbital bonding, have higher cohesive energies (i.e. they are more weakly bonded) than all transition metals , including Cu, and are therefore useful as high-energy anode metals. [9]

The difference between the metals’ ionization energies in water [9] is the other energetic contribution that can drive the reaction in a galvanic cell; it is not important in the Daniell cell because the energies of hydrated Cu2+ and Zn2+ ions happen to be similar. [9] Both atom transfer, e.g. of zinc from the metal electrode into the solution, and electron transfer, from metal atoms or to metal ions, play important roles in a galvanic cell. Concentration cells , whose electrodes and ions are made of the same metal and which are driven by an entropy increase and free-energy decrease as ion concentrations equalize, show that the electronegativity difference of the metals is not the driving force of electrochemical processes.

Galvanic cells and batteries are typically used as a source of electrical power. The energy derives from a high-cohesive-energy metal dissolving while to a lower-energy metal is deposited, and/or from high-energy metal ions plating out while lower-energy ions go into solution. Quantitatively, the electrical energy produced by a galvanic cell is approximately equal to the standard free-energy difference of the reactants and products, denoted as ΔrGo. In the Daniell cell, most of the electrical energy of ΔrGo = -213 kJ/mol can be attributed to the -207 kJ/mol difference between Zn and Cu lattice cohesive energies. [9]

## Half reactions and conventions[ edit ]

A half-cell contains a metal in two oxidation states . Inside an isolated half-cell, there is an oxidation-reduction (redox) reaction that is in chemical equilibrium , a condition written symbolically as follows (here, “M” represents a metal cation, an atom that has a charge imbalance due to the loss of “n” electrons):

Mn+ (oxidized species) + ne ⇌ M (reduced species)

A galvanic cell consists of two half-cells, such that the electrode of one half-cell is composed of metal A, and the electrode of the other half-cell is composed of metal B; the redox reactions for the two separate half-cells are thus:

An+ + ne ⇌ A
Bm+ + me ⇌ B

The overall balanced reaction is

m A + n Bm+n B + m An+

In other words, the metal atoms of one half-cell are oxidized while the metal cations of the other half-cell are reduced. By separating the metals in two half-cells, their reaction can be controlled in a way that forces transfer of electrons through the external circuit where they can do useful work .

• The electrodes are connected with a metal wire in order to conduct the electrons that participate in the reaction.
In one half-cell, dissolved metal-B cations combine with the free electrons that are available at the interface between the solution and the metal-B electrode; these cations are thereby neutralized, causing them to precipitate from solution as deposits on the metal-B electrode, a process known as plating .
This reduction reaction causes the free electrons throughout the metal-B electrode, the wire, and the metal-A electrode to be pulled into the metal-B electrode. Consequently, electrons are wrestled away from some of the atoms of the metal-A electrode, as though the metal-B cations were reacting directly with them; those metal-A atoms become cations that dissolve into the surrounding solution.
As this reaction continues, the half-cell with the metal-A electrode develops a positively charged solution (because the metal-A cations dissolve into it), while the other half-cell develops a negatively charged solution (because the metal-B cations precipitate out of it, leaving behind the anions); unabated, this imbalance in charge would stop the reaction. The solutions of the half-cells are connected by a salt bridge or a porous plate that allows ions to pass from one solution to the other, which balances the charges of the solutions and allows the reaction to continue.

By definition:

• The anode is the electrode where oxidation (loss of electrons) takes place (metal-A electrode); in a galvanic cell, it is the negative electrode, because when oxidation occurs, electrons are left behind on the electrode. [10] These electrons then flow through the external circuit to the cathode (positive electrode) (while in electrolysis, an electric current drives electron flow in the opposite direction and the anode is the positive electrode).
• The cathode is the electrode where reduction (gain of electrons) takes place (metal-B electrode); in a galvanic cell, it is the positive electrode, as ions get reduced by taking up electrons from the electrode and plate out (while in electrolysis, the cathode is the negative terminal and attracts positive ions from the solution). In both cases, the statement ‘the cathode attracts cations’ is true.

Galvanic cells, by their nature, produce direct current . The Weston cell has an anode composed of cadmium mercury amalgam , and a cathode composed of pure mercury. The electrolyte is a (saturated) solution of cadmium sulfate . The depolarizer is a paste of mercurous sulfate. When the electrolyte solution is saturated, the voltage of the cell is very reproducible; hence, in 1911, it was adopted as an international standard for voltage.

A battery is a set of galvanic cells that are connected together to form a single source of voltage. For instance, a typical 12V lead–acid battery has six galvanic cells connected in series with the anodes composed of lead and cathodes composed of lead dioxide, both immersed in sulfuric acid. Large battery rooms , for instance in a telephone exchange providing central office power to user’s telephones, may have cells connected in both series and parallel.

## Cell voltage[ edit ]

The voltage (electromotive force Eo) produced by a galvanic cell can be estimated from the standard Gibbs free energy change in the electrochemical reaction according to

$\displaystyle E_cell^o=-\Delta _rG^o/(\nu _eF)$

E

c
e
l
l

o

=

Δ

r

G

o

/

(

ν

e

F
)

\displaystyle E_cell^o=-\Delta _rG^o/(\nu _eF)

where νe is the number of electrons transferred in the balanced half reactions, and F is Faraday’s constant . However, it can be determined more conveniently by the use of a standard potential table for the two half cells involved. The first step is to identify the two metals and their ions reacting in the cell. Then one looks up the standard electrode potential ,
Eo, in volts , for each of the two half reactions . The standard potential of the cell is equal to the more positive Eo value minus the more negative Eo value.

For example, in the figure above the solutions are CuSO4 and ZnSO4. Each solution has a corresponding metal strip in it, and a salt bridge or porous disk connecting the two solutions and allowing SO2−
4
ions to flow freely between the copper and zinc solutions. To calculate the standard potential one looks up copper and zinc’s half reactions and finds:

Cu2+ + 2
e
⇌ Cu   Eo = +0.34 V
Zn2+ + 2
e
⇌ Zn   Eo = −0.76 V

Thus the overall reaction is:

Cu2+ + Zn ⇌ Cu + Zn2+

The standard potential for the reaction is then +0.34 V − (−0.76 V) = 1.10 V. The polarity of the cell is determined as follows. Zinc metal is more strongly reducing than copper metal because the standard (reduction) potential for zinc is more negative than that of copper. Thus, zinc metal will lose electrons to copper ions and develop a positive electrical charge. The equilibrium constant , K, for the cell is given by

$\displaystyle \ln K=\frac \nu _eFE_cell^oRT$

ln

K
=

ν

e

F

E

c
e
l
l

o

R
T

\displaystyle \ln K=\frac \nu _eFE_cell^oRT

where F is the Faraday constant , R is the gas constant and T is the temperature in kelvins . For the Daniell cell K is approximately equal to 1.5×1037. Thus, at equilibrium, a few electrons are transferred, enough to cause the electrodes to be charged. [11]

Actual half-cell potentials must be calculated by using the Nernst equation as the solutes are unlikely to be in their standard states,

$\displaystyle E_\texthalf-cell=E^o-\frac RT\nu _eF\ln _eQ$

E

half-cell

=

E

o

R
T

ν

e

F

ln

e

Q

\displaystyle E_\texthalf-cell=E^o-\frac RT\nu _eF\ln _eQ

where Q is the reaction quotient . When the charges of the ions in the reaction are equal, this simplifies to

$\displaystyle E_\texthalf-cell=E^o-2.303\frac RT\nu _eF\log _10\left\\textM^n+\right$/extract_itex] E half-cell = E o 2.303 R T ν e F log 10 M n + \displaystyle E_\texthalf-cell=E^o-2.303\frac RT\nu _eF\log _10\left\\textM^n+\right\ where Mn+ is the activity of the metal ion in solution. In practice concentration in mol/L is used in place of activity. The metal electrode is in its standard state so by definition has unit activity. The potential of the whole cell is obtained as the difference between the potentials for the two half-cells, so it depends on the concentrations of both dissolved metal ions. If the concentrations are the same, $\displaystyle E_cell=E_cell^o$ E c e l l = E c e l l o \displaystyle E_cell=E_cell^o and the Nernst equation is not needed under the conditions assumed here. The value of 2.303R/F is 1.9845×10−4 V/K, so at 25 °C (298.15 K) the half-cell potential will change by only 0.05918 V/νe if the concentration of a metal ion is increased or decreased by a factor of 10. $\displaystyle E_\texthalf-cell=E^o-\frac 0.05918\ \textV\nu _e\log _10\left[\textM^n+\right]$ E half-cell = E o 0.05918 V ν e log 10 [ M n + ] \displaystyle E_\texthalf-cell=E^o-\frac 0.05918\ \textV\nu _e\log _10\left[\textM^n+\right] These calculations are based on the assumption that all chemical reactions are in equilibrium. When a current flows in the circuit, equilibrium conditions are not achieved and the cell voltage will usually be reduced by various mechanisms, such as the development of overpotentials . [12] Also, since chemical reactions occur when the cell is producing power, the electrolyte concentrations change and the cell voltage is reduced. A consequence of the temperature dependency of standard potentials is that the voltage produced by a galvanic cell is also temperature dependent. ## Galvanic corrosion[ edit ] Main article: Galvanic corrosion Galvanic corrosion is a process that degrades metals electrochemically . This corrosion occurs when two dissimilar metals are in contact with each other in the presence of an electrolyte , such as salt water, forming a galvanic cell with H2 formation on the more noble metal. The resulting electrochemical potential then develops an electric current that electrolytically dissolves the less noble material. A concentration cell can be formed if the same metal is exposed to two different concentrations of electrolyte. ## Cell types[ edit ] • Concentration cell • Electrolytic cell • Electrochemical cell • Lemon battery • Thermogalvanic cell ## See also[ edit ] • Bioelectrochemical reactor • Biological cell voltage • Bio-nano generator • Cell notation • Desulfation • Electrochemical engineering • Electrode potential • Electrohydrogenesis • Electrosynthesis • Enzymatic biofuel cell • Galvanic series • Isotope electrochemistry • List of battery types • Sacrificial anode ## References[ edit ] 1. ^ “battery” (def. 4b) , Merriam-Webster Online Dictionary (2008). Retrieved 6 August 2008. 2. ^ Keithley, Joseph F (1999). Daniell Cell. John Wiley and Sons. pp. 49–51. ISBN 0-7803-1193-0 . 3. ^ Kipnis, Nahum (2003) “Changing a theory: the case of Volta’s contact electricity” , Nuova Voltiana, Vol. 5. Università degli studi di Pavia, 2003 ISBN 88-203-3273-6 . pp. 144–146 4. ^ Clarke, Edwin; Jacyna, L. S. (1992) Nineteenth-Century Origins of Neuroscientific Concepts , University of California Press. ISBN 0-520-07879-9 . p. 199 5. ^ “Milestones:Volta’s Electrical Battery Invention, 1799” . IEEE Global History Network. IEEE. Retrieved 26 July 2011. 6. ^ Ostwald, Wilhelm (1896, Veit & Co., Leipzig). Electrochemistry: History and Theory. translated from German, New Delhi: Amerind Publishing Co.; Springfield, Va, 1980, 2 vols. Check date values in: |year= ( help ) 7. ^ Haughton, Brian (2007) Hidden History: Lost Civilizations, Secret Knowledge, and Ancient Mysteries. Career Press. ISBN 1564148971 . pp. 129–132 8. ^ Ashcroft, N. W.; Mermin (1976). Solid State Physics. N. D. Brooks/Cole, Belmont, CA. 9. ^ a b c d e Schmidt-Rohr, K. (2018). “How Batteries Store and Release Energy: Explaining Basic Electrochemistry” “J. Chem. Educ.” 95: 1801-1810 http://dx.doi.org/10.1021/acs.jchemed.8b00479 10. ^ “An introduction to redox equilibria” . Chemguide. Retrieved 20 July 2014. 11. ^ Atkins, P; de Paula (2006). Physical Chemistry. J. (8th. ed.). Oxford University Press. ISBN 978-0-19-870072-2 . Chapter 7, sections on “Equilibrium electrochemistry” 12. ^ Atkins, P; de Paula (2006). Physical Chemistry. J. (8th. ed.). Oxford University Press. ISBN 978-0-19-870072-2 . Section 25.12 “Working Galvanic cells” ## External links[ edit ] • How to build a galvanic cell battery from MiniScience.com • Galvanic Cell , an animation • Interactive animation of Galvanic Cell . Chemical Education Research Group, Iowa State University. • Electron transfer reactions and redox potentials in GALVANIc cells – what happens to the ions at the phase boundary (NERNST, FARADAY) (Video by SciFox on * TIB AV-Portal ) • v • t • e Galvanic cells Types • Voltaic pile • Battery • Flow battery • Trough battery • Concentration cell • Fuel cell • Thermogalvanic cell Primary cell (non-rechargeable) • Alkaline • Aluminium–air • Bunsen • Chromic acid • Clark • Daniell • Dry • Edison-Lalande • Grove • Leclanché • Lithium • Mercury • Nickel oxyhydroxide • Silicon–air • Silver oxide • Weston • Zamboni • Zinc–air • Zinc–carbon Secondary cell (rechargeable) • Automotive • Lead–acid • gel / VRLA • Lithium–air • Lithium–ion • Lithium polymer • Lithium iron phosphate • Lithium titanate • Lithium–sulfur • Dual carbon battery • Molten salt • Nanopore • Nanowire • Nickel–cadmium • Nickel–hydrogen • Nickel–iron • Nickel–lithium • Nickel–metal hydride • Nickel–zinc • Polysulfide bromide • Potassium-ion • Rechargeable alkaline • Sodium-ion • Sodium–sulfur • Vanadium redox • Zinc–bromine • Zinc–cerium Cell parts • Anode • Binder • Catalyst • Cathode • Electrode • Electrolyte • Half-cell • Ions • Salt bridge • Semipermeable membrane Retrieved from ” https://en.wikipedia.org/w/index.php?title=Galvanic_cell&oldid=871605812 ” Categories : • Galvanic cells • Corrosion Hidden categories: • CS1 errors: dates ## Navigation menu ### Personal tools • Not logged in • Talk • Contributions • Create account • Log in ### Namespaces • Article • Talk ### Variants ### Views • Read • Edit • View history ### More ### Search ### Navigation • Main page • Contents • Featured content • Current events • Random article • Donate to Wikipedia • Wikipedia store ### Interaction • Help • About Wikipedia • Community portal • Recent changes • Contact page ### Tools • What links here • Related changes • Upload file • Special pages • Permanent link • Page information • Wikidata item • Cite this page ### Print/export • Create a book • Download as PDF • Printable version ### In other projects • Wikimedia Commons ### Languages • العربية • বাংলা • Башҡортса • Беларуская • Български • Català • Čeština • Dansk • Davvisámegiella • Deutsch • Eesti • Ελληνικά • Español • Esperanto • Euskara • فارسی • Հայերեն • हिन्दी • Hrvatski • Bahasa Indonesia • Italiano • Кыргызча • Lietuvių • Magyar • Македонски • മലയാളം • Nederlands • 日本語 • Norsk • Norsk nynorsk • ਪੰਜਾਬੀ • پنجابی • Polski • Română • Русский • සිංහල • Simple English • Slovenčina • Slovenščina • کوردی • Српски / srpski • Srpskohrvatski / српскохрватски • Suomi • Svenska • ไทย • Türkçe • Українська • Tiếng Việt • 粵語 • 中文 Edit links • This page was last edited on 2 December 2018, at 06:42 (UTC). • Text is available under the Creative Commons Attribution-ShareAlike License ; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy . Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc. , a non-profit organization. • Privacy policy • About Wikipedia • Disclaimers • Contact Wikipedia • Developers • Cookie statement • Mobile view Skip to main content # Voltaic Cells 1. Last updated 2. Save as PDF 3. Share 1. Share 2. Share 3. Tweet 4. Share • Page ID 285 • In redox reactions, electrons are transferred from one species to another. If the reaction is spontaneous, energy is released, which can then be used to do useful work. To harness this energy, the reaction must be split into two separate half reactions: the oxidation and reduction reactions. The reactions are put into two different containers and a wire is used to drive the electrons from one side to the other. In doing so, a Voltaic/ Galvanic Cell is created. ### Introduction When a redox reaction takes place, electrons are transferred from one species to the other. If the reaction is spontaneous, energy is released, which can be used to do work. Consider the reaction of a solid copper (Cu(s)) in a silver nitrate solution (AgNO3(s)). \[2Ag^+_(aq) + Cu_(s) \leftrightharpoons Cu^2+_(aq) + 2Ag_(s)$

The $$AgNO_3\;(s)$$ dissociates in water to produce $$Ag^+_(aq)$$ ions and $$NO­­^-_3\;(aq)$$ ions. The NO3(aq) ions can be ignored since they are spectator ions and do not participate in the reaction. In this reaction, a copper electrode is placed into a solution containing silver ions. The Ag+(aq) will readily oxidize Cu(s) resulting in Cu2+(aq), while reducing itself to Ag(s).

This reaction releases energy. When the copper electrode solid is placed directly into a silver nitrate solution, however, the energy is lost as heat and cannot be used to do work. In order to harness this energy and use it do useful work, we must split the reaction into two separate half reactions; The oxidation and reduction reactions. A wire connects the two reactions and allows electrons to flow from one side to the other. In doing so, we have created a Voltaic/ Galvanic Cell.

Figure $$\PageIndex1$$: Voltaic Cell

A Voltaic Cell (also known as a 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 (a strip of metal, M) within a solution containing Mn+ ions in which M is any arbitrary metal. The two half cells are linked together by a wire running from one electrode to the other. A salt bridge also connects to the half cells. The functions of these parts are discussed below.

### Half Cells

Half of the redox reaction occurs at each half cell. Therefore, we can say that in each half-cell a half-reaction is taking place. When the two halves are linked together with a wire and a salt bridge, an electrochemical cell is created.

### Electrodes

An electrode is strip of metal on which the reaction takes place. In a voltaic cell, the oxidation and reduction of metals occurs at the electrodes. There are two electrodes in a voltaic cell, one in each half-cell. The cathode is where reduction takes place and oxidation takes place at the anode.

Through electrochemistry, these reactions are reacting upon metal surfaces, or electrodes. An oxidation-reduction equilibrium is established between the metal and the substances in solution. When electrodes are immersed in a solution containing ions of the same metal, it is called a half-cell. Electrolytes are ions in solution, usually fluid, that conducts electricity through ionic conduction. Two possible interactions can occur between the metal atoms on the electrode and the ion solutions.

1. Metal ion Mn+ from the solution may collide with the electrode, gaining "n" electrons from it, and convert to metal atoms. This means that the ions are reduced.
2. Metal atom on the surface may lose "n" electrons to the electrode and enter the solution as the ion Mn+ meaning that the metal atoms are oxidized.

When an electrode is oxidized in a solution, it is called an anode and when an electrode is reduced in solution. it is called a cathode.

• Anode: The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. In the reaction above, the anode is the Cu(s) since it increases in oxidation state from 0 to +2.
• Cathode: The cathode is where the reduction reaction takes place. This is where the metal electrode gains electrons. Referring back to the equation above, the cathode is the Ag(s) as it decreases in oxidation state from +1 to 0.

Remembering Oxidation and Reduction

When it comes to redox reactions, it is important to understand what it means for a metal to be “oxidized” or “reduced”. An easy way to do this is to remember the phrase “OIL RIG”.

OIL = Oxidization is Loss (of e)

RIG = Reduction is Gain (of e)

In the case of the example above $$Ag^+_(aq)$$ gains an electron meaning it is reduced. $$Cu_(s)$$ loses two electrons thus it is oxidized.

The salt bridge is a vital component of any voltaic cell. It is a tube filled with an electrolyte solution such as KNO3(s) or KCl(s). The purpose of the salt bridge is to keep the solutions electrically neutral and allow the free flow of ions from one cell to another. Without the salt bridge, positive and negative charges will build up around the electrodes causing the reaction to stop.

The purpose of the salt bridge is to keep the solutions electrically neutral and allow the free flow of ions from one cell to another.

### Flow of Electrons

Electrons always flow from the anode to the cathode or from the oxidation half cell to the reduction half cell. In terms of Eocell of the half reactions, the electrons will flow from the more negative half reaction to the more positive half reaction. A cell diagram is a representation of an electrochemical cell. The figure below illustrates a cell diagram for the voltaic shown in Figure $$\PageIndex1$$ above.

Figure $$\PageIndex2$$: Cell Diagram. The figure below illustrates a cell diagram for the voltaic shown in Figure $$\PageIndex1$$.

When drawing a cell diagram, we follow the following conventions. The anode is always placed on the left side, and the cathode is placed on the right side. The salt bridge is represented by double vertical lines (||). The difference in the phase of an element is represented by a single vertical line (|), while changes in oxidation states are represented by commas (,).

When asked to construct a cell diagram follow these simple instructions. Consider the following reaction:

$2Ag^+_(aq) + Cu_(s) \rightleftharpoons Cu^2+_(aq) + 2Ag_(s)$

Step 1: Write the two half-reactions.

$Ag^+_(aq) + e^- \rightleftharpoons Ag_(s)$

$Cu_(s) \rightleftharpoons Cu^2+_(aq) + 2e^-$

Step 2: Identify the cathode and anode.

$$Cu_(s)$$ is losing electrons thus being oxidized; oxidation occurs at the anode.

• Anode (where oxidation occurs): $$Cu_(s) \rightleftharpoons Cu^2+_(aq) + 2e^-$$

$$Ag^+$$ is gaining electrons thus is being reduced; reduction happens at the cathode.

• Cathode (where reduction occurs): $$Ag^+_(aq) + e^- \rightleftharpoons Ag_(s)$$

Step 3: Construct the Cell Diagram.

$Cu_(s) | Cu^2+_(aq) || Ag^+_(aq) | Ag_(s)$

The anode always goes on the left and cathode on the right. Separate changes in phase by | and indicate the the salt bridge with ||. The lack of concentrations indicates solutions are under standard conditions (i.e., 1 M)

Example $$\PageIndex1$$

Consider the following two reactions:

$Cu^2+_(aq) + Ba_(s) \rightarrow Cu_(s) + Ba^2+_(aq)$

$2Al_(s) + 3Sn^2+_(aq) \rightarrow 2Al^3+_(aq) + 3Sn_(s)$

1. Split the reaction into half reactions and determine their standard reduction potentials. Indicate which would be the anode and cathode.
2. Construct a cell diagram for the following each reactions.
3. Determine the $$E^o_cell$$ for the voltaic cell formed by each reaction.

Solution

1.a) Ba2+(aq) → Ba(s) + 2e- with SRP (for opposite reaction) Eo = -2.92 V (Anode; where oxidation happens)

Cu2+(aq) + 2e- → Cu(s) with SRP Eo = +0.340 V (Cathode; where reduction happens)

1.b) Al3+(aq) → Al(s) + 3e with SRP (for opposite reaction) Eo = -1.66 V (Anode; where oxidation happens)

Sn2+(aq) +2e– → Sn(s)  with SRP Eo = -0.137 V (Cathode; where reduction happens)

2.a) Ba2+(aq) | Ba(s) || Cu(s) | Cu2+(aq)

2.b) Al(s) | Al3+(aq) || Sn2+(aq) | Sn(s)

3.a) Eocell = 0.34 – (-2.92) = 3.26 V

3.b) Eocell = -0.137 – (-1.66) = 1.523 V

### Cell Voltage/Cell Potential

The readings from the voltmeter give the reaction’s cell voltage or potential difference between it’s two two half-cells. Cell voltage is also known as cell potential or electromotive force (emf) and it is shown as the symbol $$E_cell$$.

Standard Cell Potential:

$E^o_cell = E^o_cathode – E^o_anode$

The Eo values 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 electrochemical cell. If Eocell is positive the reaction is spontaneous and it is a voltaic cell. If the Eocell is negative, the reaction is non-spontaneous and it is referred to as an electrolytic cell.

### References

1. Brady, James E., Holum, John R. “Chemistry: The Study of Matter and Its Changes”, John Wiley & Sons Inc 1993
2. Brown, Theodore L., LeMay, H. Eugene Jr. “Chemistry: The Central Science” Third Edition, Prentice-Hall, Inc. Englewood Cliffs, N.J. 07632 1985
3. Brown, Theodore L., LeMay, H. Eugene Jr., Bursten, Bruce E. “Chemistry: The Central Science” Fifth Edition, Prentice-Hall, Inc. Englewood Cliffs, N.J. 07632 1991
4. Gesser, Hyman D. “ Descriptive Principles of Chemistry”, C.V. Mosby Company 1974
5. Harwood, William, Herring, Geoffrey, Madura, Jeffry, and Petrucci, Ralph, General Chemistry: Principles and Modern Applications, Ninth Edition, Upper Saddle River,New Jersey, Pearson Prentice Hall, 2007.
6. Petrucci, Ralph H. Genereal Chemistry: Principles and Modern Applications 9th Ed. New Jersey: Pearson Education Inc. 2007.
7. Vassos Basil H. Electroanaytical Chemistry. New York: Wiley-Interscience Publication. 1983.
8. Zumdahl, Steven S. Chemistry 7th Ed. New York: Houghton Mifflin Company. 2007.

### Contributors

• Shamsher Singh, Deborah Gho

# Voltaic Cells

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• Page ID
285
• In redox reactions, electrons are transferred from one species to another. If the reaction is spontaneous, energy is released, which can then be used to do useful work. To harness this energy, the reaction must be split into two separate half reactions: the oxidation and reduction reactions. The reactions are put into two different containers and a wire is used to drive the electrons from one side to the other. In doing so, a Voltaic/ Galvanic Cell is created.

### Introduction

When a redox reaction takes place, electrons are transferred from one species to the other. If the reaction is spontaneous, energy is released, which can be used to do work. Consider the reaction of a solid copper (Cu(s)) in a silver nitrate solution (AgNO3(s)).

$2Ag^+_(aq) + Cu_(s) \leftrightharpoons Cu^2+_(aq) + 2Ag_(s)$

The $$AgNO_3\;(s)$$ dissociates in water to produce $$Ag^+_(aq)$$ ions and $$NO­­^-_3\;(aq)$$ ions. The NO3(aq) ions can be ignored since they are spectator ions and do not participate in the reaction. In this reaction, a copper electrode is placed into a solution containing silver ions. The Ag+(aq) will readily oxidize Cu(s) resulting in Cu2+(aq), while reducing itself to Ag(s).

This reaction releases energy. When the copper electrode solid is placed directly into a silver nitrate solution, however, the energy is lost as heat and cannot be used to do work. In order to harness this energy and use it do useful work, we must split the reaction into two separate half reactions; The oxidation and reduction reactions. A wire connects the two reactions and allows electrons to flow from one side to the other. In doing so, we have created a Voltaic/ Galvanic Cell.

Figure $$\PageIndex1$$: Voltaic Cell

A Voltaic Cell (also known as a 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 (a strip of metal, M) within a solution containing Mn+ ions in which M is any arbitrary metal. The two half cells are linked together by a wire running from one electrode to the other. A salt bridge also connects to the half cells. The functions of these parts are discussed below.

### Half Cells

Half of the redox reaction occurs at each half cell. Therefore, we can say that in each half-cell a half-reaction is taking place. When the two halves are linked together with a wire and a salt bridge, an electrochemical cell is created.

### Electrodes

An electrode is strip of metal on which the reaction takes place. In a voltaic cell, the oxidation and reduction of metals occurs at the electrodes. There are two electrodes in a voltaic cell, one in each half-cell. The cathode is where reduction takes place and oxidation takes place at the anode.

Through electrochemistry, these reactions are reacting upon metal surfaces, or electrodes. An oxidation-reduction equilibrium is established between the metal and the substances in solution. When electrodes are immersed in a solution containing ions of the same metal, it is called a half-cell. Electrolytes are ions in solution, usually fluid, that conducts electricity through ionic conduction. Two possible interactions can occur between the metal atoms on the electrode and the ion solutions.

1. Metal ion Mn+ from the solution may collide with the electrode, gaining "n" electrons from it, and convert to metal atoms. This means that the ions are reduced.
2. Metal atom on the surface may lose "n" electrons to the electrode and enter the solution as the ion Mn+ meaning that the metal atoms are oxidized.

When an electrode is oxidized in a solution, it is called an anode and when an electrode is reduced in solution. it is called a cathode.

• Anode: The anode is where the oxidation reaction takes place. In other words, this is where the metal loses electrons. In the reaction above, the anode is the Cu(s) since it increases in oxidation state from 0 to +2.
• Cathode: The cathode is where the reduction reaction takes place. This is where the metal electrode gains electrons. Referring back to the equation above, the cathode is the Ag(s) as it decreases in oxidation state from +1 to 0.

Remembering Oxidation and Reduction

When it comes to redox reactions, it is important to understand what it means for a metal to be “oxidized” or “reduced”. An easy way to do this is to remember the phrase “OIL RIG”.

OIL = Oxidization is Loss (of e)

RIG = Reduction is Gain (of e)

In the case of the example above $$Ag^+_(aq)$$ gains an electron meaning it is reduced. $$Cu_(s)$$ loses two electrons thus it is oxidized.

The salt bridge is a vital component of any voltaic cell. It is a tube filled with an electrolyte solution such as KNO3(s) or KCl(s). The purpose of the salt bridge is to keep the solutions electrically neutral and allow the free flow of ions from one cell to another. Without the salt bridge, positive and negative charges will build up around the electrodes causing the reaction to stop.

The purpose of the salt bridge is to keep the solutions electrically neutral and allow the free flow of ions from one cell to another.

### Flow of Electrons

Electrons always flow from the anode to the cathode or from the oxidation half cell to the reduction half cell. In terms of Eocell of the half reactions, the electrons will flow from the more negative half reaction to the more positive half reaction. A cell diagram is a representation of an electrochemical cell. The figure below illustrates a cell diagram for the voltaic shown in Figure $$\PageIndex1$$ above.

Figure $$\PageIndex2$$: Cell Diagram. The figure below illustrates a cell diagram for the voltaic shown in Figure $$\PageIndex1$$.

When drawing a cell diagram, we follow the following conventions. The anode is always placed on the left side, and the cathode is placed on the right side. The salt bridge is represented by double vertical lines (||). The difference in the phase of an element is represented by a single vertical line (|), while changes in oxidation states are represented by commas (,).

When asked to construct a cell diagram follow these simple instructions. Consider the following reaction:

$2Ag^+_(aq) + Cu_(s) \rightleftharpoons Cu^2+_(aq) + 2Ag_(s)$

Step 1: Write the two half-reactions.

$Ag^+_(aq) + e^- \rightleftharpoons Ag_(s)$

$Cu_(s) \rightleftharpoons Cu^2+_(aq) + 2e^-$

Step 2: Identify the cathode and anode.

$$Cu_(s)$$ is losing electrons thus being oxidized; oxidation occurs at the anode.

• Anode (where oxidation occurs): $$Cu_(s) \rightleftharpoons Cu^2+_(aq) + 2e^-$$

$$Ag^+$$ is gaining electrons thus is being reduced; reduction happens at the cathode.

• Cathode (where reduction occurs): $$Ag^+_(aq) + e^- \rightleftharpoons Ag_(s)$$

Step 3: Construct the Cell Diagram.

$Cu_(s) | Cu^2+_(aq) || Ag^+_(aq) | Ag_(s)$

The anode always goes on the left and cathode on the right. Separate changes in phase by | and indicate the the salt bridge with ||. The lack of concentrations indicates solutions are under standard conditions (i.e., 1 M)

Example $$\PageIndex1$$

Consider the following two reactions:

$Cu^2+_(aq) + Ba_(s) \rightarrow Cu_(s) + Ba^2+_(aq)$

$2Al_(s) + 3Sn^2+_(aq) \rightarrow 2Al^3+_(aq) + 3Sn_(s)$

1. Split the reaction into half reactions and determine their standard reduction potentials. Indicate which would be the anode and cathode.
2. Construct a cell diagram for the following each reactions.
3. Determine the $$E^o_cell$$ for the voltaic cell formed by each reaction.

Solution

1.a) Ba2+(aq) → Ba(s) + 2e- with SRP (for opposite reaction) Eo = -2.92 V (Anode; where oxidation happens)

Cu2+(aq) + 2e- → Cu(s) with SRP Eo = +0.340 V (Cathode; where reduction happens)

1.b) Al3+(aq) → Al(s) + 3e with SRP (for opposite reaction) Eo = -1.66 V (Anode; where oxidation happens)

Sn2+(aq) +2e– → Sn(s)  with SRP Eo = -0.137 V (Cathode; where reduction happens)

2.a) Ba2+(aq) | Ba(s) || Cu(s) | Cu2+(aq)

2.b) Al(s) | Al3+(aq) || Sn2+(aq) | Sn(s)

3.a) Eocell = 0.34 – (-2.92) = 3.26 V

3.b) Eocell = -0.137 – (-1.66) = 1.523 V

### Cell Voltage/Cell Potential

The readings from the voltmeter give the reaction’s cell voltage or potential difference between it’s two two half-cells. Cell voltage is also known as cell potential or electromotive force (emf) and it is shown as the symbol $$E_cell$$.

Standard Cell Potential:

$E^o_cell = E^o_cathode – E^o_anode$

The Eo values 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 electrochemical cell. If Eocell is positive the reaction is spontaneous and it is a voltaic cell. If the Eocell is negative, the reaction is non-spontaneous and it is referred to as an electrolytic cell.

### References

1. Brady, James E., Holum, John R. “Chemistry: The Study of Matter and Its Changes”, John Wiley & Sons Inc 1993
2. Brown, Theodore L., LeMay, H. Eugene Jr. “Chemistry: The Central Science” Third Edition, Prentice-Hall, Inc. Englewood Cliffs, N.J. 07632 1985
3. Brown, Theodore L., LeMay, H. Eugene Jr., Bursten, Bruce E. “Chemistry: The Central Science” Fifth Edition, Prentice-Hall, Inc. Englewood Cliffs, N.J. 07632 1991
4. Gesser, Hyman D. “ Descriptive Principles of Chemistry”, C.V. Mosby Company 1974
5. Harwood, William, Herring, Geoffrey, Madura, Jeffry, and Petrucci, Ralph, General Chemistry: Principles and Modern Applications, Ninth Edition, Upper Saddle River,New Jersey, Pearson Prentice Hall, 2007.
6. Petrucci, Ralph H. Genereal Chemistry: Principles and Modern Applications 9th Ed. New Jersey: Pearson Education Inc. 2007.
7. Vassos Basil H. Electroanaytical Chemistry. New York: Wiley-Interscience Publication. 1983.
8. Zumdahl, Steven S. Chemistry 7th Ed. New York: Houghton Mifflin Company. 2007.

### Contributors

• Shamsher Singh, Deborah Gho