23.5: Galvanic Cells
Lesson Objectives
The student will:
 describe the conditions necessary for a cell to be a standard cell.
 balance the redox equation, calculate the standard cell potential, and determine the direction of electron flow in the external circuit given a table of standard reduction potentials and a diagram or description of a Galvanic cell.
Vocabulary
 electrochemical cell
 salt bridge
 voltage
Introduction
Electrochemistry can be defined as the study of the interchange of electrical and chemical energy. Electrochemical reactions generate an electric current from a spontaneous chemical reaction.
Spontaneous Redox Reactions
Electrolysis requires a direct electric current in order to drive a nonspontaneous redox reaction, but spontaneous redox reactions are also very common. These reactions occur because the products contain less potential energy in their bonds than the reactants do. The energy produced from excess potential energy not only allows the reaction to occur, but it often gives off energy to the surroundings. Some of these reactions can be physically arranged so that the energy given off is in the form of an electric current. These are the type of reactions that occur inside batteries. When a reaction is arranged to produce an electric current as it runs, the arrangement is called an electrochemical cell or a Galvanic Cell.
If a strip of copper is placed in a solution of silver nitrate, the following reaction takes place:


 \begin{align*}2 \ \mathrm{Ag}^+ + \mathrm{Cu} \rightarrow 2 \ \mathrm{Ag} + \mathrm{Cu}^{2+}\end{align*}

In this reaction (illustrated below), copper atoms are donating electrons to silver ions so that the silver ions are reduced to silver atoms and the copper atoms are oxidized to copper(II) ions.
As the reaction occurs, an observer would see the solution slowly turn blue (\begin{align*}\mathrm{Cu}^{2+}\end{align*} ions are blue in solution), and a mass of solid silver atoms would build up on the copper strip.
Electrochemical Cells
The reaction \begin{align*}2 \ \mathrm{Ag}^+ + \mathrm{Cu} \rightarrow 2 \ \mathrm{Ag} + \mathrm{Cu}^{2+}\end{align*} is one that could be physically arranged to produce an external electric current. To do this, the two halfreactions must occur in separate compartments, and the separate compartments must remain in contact through an ionic solution and an external wire.
In the electrochemical cell illustrated above, the copper metal must be separated from the silver ions to avoid a direct reaction. Each electrode in its solution could be represented by the following halfreactions:


 \begin{align*}\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2 \ e^\end{align*}



 \begin{align*}\mathrm{Ag} \rightarrow \mathrm{Ag}^+ + e^\end{align*}

The external wire will allow electrons to flow between the metal strips. In each halfcell, atoms may be oxidized to ions, leaving excess electrons on the electrode. It can be determined experimentally that electrons will flow in the wire from the copper electrode to the silver electrode. The reason for the direction of electron flow can be explained in several ways. We could say that silver ions have a greater electron affinity than copper ions, so the silver atoms pull the electrons through the wire from copper. On the other hand, we could also say that copper atoms have a greater tendency to give up electrons than silver, so the copper electrode becomes more negative and will push the electrons through the wire toward the silver electrode. Whichever way we look at it, the electrons flow from copper to silver in the external wire. The silver electrode, therefore, will acquire an excess of electrons that causes the silver halfreaction to run in the reverse direction.


 \begin{align*}\mathrm{Ag}^+ + e^ \rightarrow \mathrm{Ag}\end{align*}

As the electrons on the silver electrode are used up in the reduction of silver ions, more copper atoms are oxidized, and more electrons sent through the wire. The net reaction for the entire cell is:


 \begin{align*}2 \ \mathrm{Ag}^+ _{(aq)} + \mathrm{Cu}_{(s)} \rightarrow 2 \ \mathrm{Ag}_{(aq)} + \mathrm{Cu}^{2+}_{(aq)}\end{align*}

As the cell runs, electrons are transferred from the copper halfcell to the silver halfcell. Unchecked, this reaction would result in the silver halfcell becoming negatively charged and the copper halfcell becoming positively charged. Once the halfcells became charged in that manner, the reaction could only continue to run if it produced sufficient energy to take electrons away from a positive charge and push them onto an already negatively charged halfcell. Chemical reactions, however, do not produce enough energy to push electrons against a charge gradient.
The salt bridge is present to allow negative ions to flow from the silver halfcell to the copper halfcell, and positive ions to flow from the copper halfcell to the silver halfcell. This migration of ions balances the charge movement of the electrons. The salt bridge is the upsidedown Ushaped tube connecting the two beakers in the image above. This tube is filled with an ionic solution, and the ends are fitted with porous plugs. An ionic solution is chosen such that neither of its ions will react chemically with any of the other ions in the system. The porous plugs are there to avoid general mixing but to allow ion migration. If the salt bridge were not present or were removed, the reaction would immediately stop. As long as the two beakers remain neutral and there are sufficient reactants (\begin{align*}\mathrm{Ag}^+\end{align*} and \begin{align*}\mathrm{Cu}\end{align*}) to continue the reaction, the reaction will continue to run.
We can now see that the silver electrode is the cathode because cations migrate toward it and because reduction occurs there. The copper electrode is the anode because anions migrate toward it and because oxidation occurs there. As the cell runs and produces electric current, the mass of the silver electrode increases because when a silver ion is reduced to a silver atom, it attaches to the silver electrode. The copper electrode will lose mass because as a copper atom is oxidized to a copper ion, it dissolves in the solution. Eventually, the silver electrode will show a mass of attached solid silver, and the copper electrode will develop holes and edges as evidence of the reaction wearing away the strip.
The electrons that pass through the external circuit can do useful work, such as lighting lights, running cell phones, and so forth. Several cells can be operated together to produce greater current. When we have a series of cells operating together as one, we call the arrangement a battery. If the light bulb is removed from the circuit with the electrochemical cell and replaced with a voltmeter (see illustration below), the voltmeter will measure the voltage (electrical potential energy per unit charge) of the combination of halfcells.
The size of the voltage produced by a cell depends on the temperature, the metals used for electrodes, and the concentrations of the ions in the solutions. If you increase the concentration of the reactant ion (not the product ion), the reaction rate will increase and so will the voltage.
It may seem complicated to construct an electrochemical cell because of all its complexities. Electrochemical cells, however, are actually easy to make and sometimes even occur accidentally. If you take two coins of different different metal composition, one copper and one silver for example, and push them part way through the peel of a whole lemon (as illustrated below), upon connecting the two coins with a wire, a small electric current will flow.
Electrochemical cells occasionally occur accidentally when two water pipes of different material are connected. The reactions at the joint cause a great deal of corrosion. Plumbing professionals take great care to make sure such reactions do not occur at pipe joints.
Another common way to construct an electrochemical cell is with the use of a porous cup, as shown in the figure above. A porous cup is made of a material that is permeable to ions and allows the ions to migrate through the walls. The cup is not, however, porous enough to allow liquid to flow through. Unfinished clay is one such substance. The porous cup is soaked in ionic solution to ensure that the pores are filled with ionic solution. One electrode and its solution are placed in the porous cup, and the other electrode and its corresponding ionic solution is placed in a beaker. The porous cup with its contents is then set inside the beaker, and a wire is connected between the electrodes. The walls of the porous cup act as the salt bridge in this cell.
This animation shows the function of a galvanic cell: http://www.youtube.com/watch?v=A0VUsoeT9aM (1:33).
The Standard Hydrogen HalfCell
When designing cells for specific purposes, it would be very convenient if chemists were able to predict which halfreaction would gain electrons and which would lose. It would also be useful to predict the voltages that will be produced by various combinations of halfcells. Recall, however, that both the temperature and the concentrations of the ion solutions can change the reaction rate and the voltage of a cell. Therefore, the conditions for comparing halfreactions must be standardized. The conditions chosen to be the standard for halfcells are \begin{align*}25^\circ\mathrm{C}\end{align*}, \begin{align*}1.0 \ \mathrm{M}\end{align*} for all solutions, and \begin{align*}1.0\end{align*} atmosphere pressure.
It would be convenient if a chemist could make 100 standard halfcells, attach them to a voltmeter, and measure the voltage of each standard halfcell. Unfortunately, as you know, halfcells will not run alone. There must be two halfcells connected together; one where oxidation occurs and one where reduction occurs. When you connect two halfcells together, you can indeed measure the voltage produced, but you do not know how much of the voltage is due to each of the two standard halfcells. Chemists solved the problem by assigning the standard hydrogen halfcell a potential of zero. When another standard halfcell was combined with the hydrogen halfcell, the cell voltage could be measured, and the potential of the cell assigned to the other halfcell.
For example, the standard silver halfcell, \begin{align*}\mathrm{Ag}^+ + e^ \rightarrow \mathrm{Ag}\end{align*}, could be combined with the standard hydrogen halfcell, \begin{align*}2 \ \mathrm{H}^+ + 2e^ \rightarrow \mathrm{H}_2\end{align*}, and the resultant voltage of the cell could then be measured. In this cell, the silver ions are reduced and the hydrogen atoms are oxidized. Since both half cells were written in terms of reduction, in order to write the equation for the cell, the hydrogen halfreaction must be reversed.


 \begin{align*} \begin{array}{rlll} 2 \ \mathrm{Ag}^+ + 2 \ e^ & \rightarrow & 2 \ \mathrm{Ag} & \text{(reduction)}\\ \mathrm{H}_2 & \rightarrow & 2 \ \mathrm{H}^+ + 2 \ e^ & \text{(oxidation)}\\ \hline 2 \ \mathrm{Ag}^+ + \mathrm{H}_2 & \rightarrow & 2 \ \mathrm{H}^+ + 2 \ \mathrm{Ag} & \text{(net reaction)} \end{array} \end{align*}

When this cell runs, it produces a voltage of \begin{align*}0.80 \ \mathrm{volts}\end{align*}. This voltage is assigned to the silver halfcell as its voltage when the silver ions are reduced. The assigned voltages of \begin{align*}0.80 \ \mathrm{V}\end{align*} for the silver halfcell and \begin{align*}0.00 \ \mathrm{V}\end{align*} for the hydrogen halfcell do not have any meaning in terms of voltages on an absolute scale, but they are perfectly accurate as an indication of the difference in the abilities of the two halfcells to take electrons. The silver halfcell has a greater pull on electrons than the hydrogen halfcell, and it is stronger by \begin{align*}0.80 \ \mathrm{volts}\end{align*}.
It was found that some of the standard halfcells are not strong enough to take electrons from the hydrogen halfcell. In fact, the hydrogen halfcell is the one that takes the electrons and forces the other halfreaction to oxidize. Since the hydrogen halfcell was assigned a strength of zero and these halfcells are weaker than hydrogen, their assigned voltages are negative numbers.
When many halfcells have been combined with the hydrogen halfcell and the voltages measured, we have a list of halfcells in order of their strength at taking electrons, and we have a numerical value for the difference in their strength compared to the hydrogen halfcell. Table below is an expanded list of halfreactions and their assigned voltages based on comparison to the hydrogen halfcell.
HalfReactions  \begin{align*}E^\circ \;\mathrm{(volts)}\end{align*}  

\begin{align*}\mathrm{F}_{2(g)} + 2 \ e^ \rightarrow 2 \ \mathrm{F}^\end{align*}  \begin{align*}+ 2.87\end{align*}  
\begin{align*}\mathrm{MnO}_4 ^ + 8 \ \mathrm{H}^+ + 5 \ e^ \rightarrow Mn^{2+} + 4 \ \mathrm{H}_2\mathrm{O}\end{align*}  \begin{align*}+ 1.52\end{align*}  
\begin{align*}\mathrm{Au}^{3+} + 3 \ e^ \rightarrow \mathrm{Au}_{(s)}\end{align*}  \begin{align*}+ 1.50\end{align*}  
\begin{align*}\mathrm{Cl}_{2(g)} + 2 \ e^ \rightarrow 2 \ \mathrm{Cl}^\end{align*}  \begin{align*}+ 1.36\end{align*}  
\begin{align*}\mathrm{Cr}_2\mathrm{O}_7^{2} + 14 \ \mathrm{H}^+ + 6 \ e^ \rightarrow 2 \ \mathrm{Cr}^{3+} + 7 \ \mathrm{H}_2\mathrm{O}\end{align*}  \begin{align*}+ 1.33\end{align*}  
\begin{align*}\mathrm{MnO}_{2(s)} + 4 \ \mathrm{H}^+ + 2 \ e^ \rightarrow \mathrm{Mn}^{2+} + 2 \ \mathrm{H}_2\mathrm{O}\end{align*}  \begin{align*}+ 1.28\end{align*}  
\begin{align*}\frac{1}{2} \ \mathrm{O}_{2(g)} + 2 \ \mathrm{H}^+ + 2 \ e^ \rightarrow \mathrm{H}_2\mathrm{O}\end{align*}  \begin{align*}+ 1.23\end{align*}  
\begin{align*}\mathrm{Br}_{2(l)} + 2 \ e^ \rightarrow 2 \ \mathrm{Br}^\end{align*}  \begin{align*}+ 1.06\end{align*}  
\begin{align*}\mathrm{NO}_3 ^ + 4 \ \mathrm{H}^+ + 3 \ e^ \rightarrow \mathrm{NO}_{(g)} + 2 \ \mathrm{H}_2\mathrm{O}\end{align*}  \begin{align*}+ 0.96\end{align*}  
\begin{align*}\mathrm{Ag}^+ + e^ \rightarrow \mathrm{Ag}\end{align*}  \begin{align*}+ 0.80\end{align*}  
\begin{align*}\mathrm{Hg}^{2+} + 2 \ e^ \rightarrow \mathrm{Hg}_{(l)}\end{align*}  \begin{align*}+ 0.78\end{align*}  
\begin{align*}\mathrm{NO}_3 ^ + 2 \ \mathrm{H}^+ + e^ \rightarrow \mathrm{NO}_{2(g)} + \mathrm{H}_2\mathrm{O}\end{align*}  \begin{align*}+ 0.78\end{align*}  
\begin{align*}\mathrm{Fe}^{3+} + e^ \rightarrow \mathrm{Fe}^{2+}\end{align*}  \begin{align*}+ 0.77\end{align*}  
\begin{align*}\mathrm{I}_{2(s)} + 2 \ e^ \rightarrow 2 \ \mathrm{I}^\end{align*}  \begin{align*}+ 0.53\end{align*}  
\begin{align*}\mathrm{Cu}^{2+} + 2 \ e^ \rightarrow \mathrm{Cu}_{(s)}\end{align*}  \begin{align*}+ 0.34\end{align*}  
\begin{align*}\mathrm{SO}_4^{2} + 4 \ \mathrm{H}^+ + 2 \ e^ \rightarrow \mathrm{SO}_{2(g)} + 2 \ \mathrm{H}_2\mathrm{O}\end{align*}  \begin{align*}+ 0.17\end{align*}  
\begin{align*}\mathrm{Sn}^4+ + 2 \ e^ \rightarrow \mathrm{Sn}^{2+}\end{align*}  \begin{align*}+ 0.15\end{align*}  
\begin{align*}2 \ \mathrm{H}^+ + 2 \ e^ \rightarrow \mathrm{H}_{2(g)}\end{align*}  
\begin{align*}\mathrm{Pb}^{2+} + 2 \ e^ \rightarrow \mathrm{Pb}_{(s)}\end{align*}  \begin{align*} 0.13\end{align*}  
\begin{align*}\mathrm{Sn}^{2+} + 2 \ e^ \rightarrow \mathrm{Sn}_{(s)}\end{align*}  \begin{align*} 0.14\end{align*}  
\begin{align*}\mathrm{Ni}^{2+} + 2 \ e^ \rightarrow \mathrm{Ni}_{(s)}\end{align*}  \begin{align*} 0.25\end{align*}  
\begin{align*}Co^{2+} + 2\ e^ \rightarrow Co_{(s)}\end{align*}  \begin{align*} 0.28\end{align*}  
\begin{align*}\mathrm{Fe}^{2+} + 2 \ e^ \rightarrow \mathrm{Fe}_{(s)}\end{align*}  \begin{align*} 0.44\end{align*}  
\begin{align*}\mathrm{Cr}^{3+} + 3 \ e^ \rightarrow \mathrm{Cr}_{(s)}\end{align*}  \begin{align*} 0.74\end{align*}  
\begin{align*}\mathrm{Zn}^{2+} + 2 \ e^ \rightarrow \mathrm{Zn}_{(s)}\end{align*}  \begin{align*} 0.76\end{align*}  
\begin{align*}2 \ \mathrm{H}_2\mathrm{O} + 2 \ e^ \rightarrow 2 \ \mathrm{OH}^ + \mathrm{H}_{2(g)}\end{align*}  \begin{align*} 0.83\end{align*}  
\begin{align*}\mathrm{Mn}^{2+} + 2 \ e^ \rightarrow \mathrm{Mn}_{(s)}\end{align*}  \begin{align*} 1.18\end{align*}  
\begin{align*}\mathrm{Al}^{3+} + 3 \ e^ \rightarrow \mathrm{Al}_{(s)}\end{align*}  \begin{align*} 1.66\end{align*}  
\begin{align*}\mathrm{Mg}^{2+} + 2 \ e^ \rightarrow \mathrm{Mg}_{(s)}\end{align*}  \begin{align*} 2.37\end{align*}  
\begin{align*}\mathrm{Na}^+ + e^ \rightarrow \mathrm{Na}_{(s)}\end{align*}  \begin{align*} 2.71\end{align*}  
\begin{align*}\mathrm{Ca}^{2+} + 2 \ e^ \rightarrow \mathrm{Ca}_{(s)}\end{align*}  \begin{align*} 2.87\end{align*}  
\begin{align*}\mathrm{Sr}^{2+} + 2 \ e^ \rightarrow \mathrm{Sr}_{(s)}\end{align*}  \begin{align*} 2.89\end{align*}  
\begin{align*}\mathrm{Ba}^{2+} + 2 \ e^ \rightarrow \mathrm{Ba}_{(s)}\end{align*}  \begin{align*} 2.90\end{align*}  
\begin{align*}\mathrm{K}^+ + e^ \rightarrow \mathrm{K}_{(s)}\end{align*}  \begin{align*} 2.92\end{align*}  
\begin{align*}\mathrm{Li}^+ + e^ \rightarrow \mathrm{Li}_{(s)}\end{align*}  \begin{align*} 3.00\end{align*} 
Example:
In a cell formed by the \begin{align*}\mathrm{Cu}^{2+} + 2 \ e^ \rightarrow \mathrm{Cu}\end{align*} and the \begin{align*}\mathrm{Zn}^{2+} + 2 \ e^ \rightarrow \mathrm{Zn}\end{align*} standard halfcells, determine which halfcell will undergo oxidation, which halfcell will undergo reduction, the voltage for the cell, and the balanced equation for the cell.
Solution:
Simply by looking at the placement of the halfreactions in the standard reduction potential chart, we can determine that the copper halfcell is closer to the top of the chart than the zinc halfcell, so we know that copper is better at attracting electrons and that halfreaction will run in the reduction direction. The zinc halfcell will be forced to give up electrons and be oxidized, so that halfreaction will also need to be reversed.


 \begin{align*} \begin{array}{rlll} \mathrm{Cu}^{2+} + 2 \ e^ & \rightarrow & \mathrm{Cu}_{(s)} & \text{(reduction)}\\ \mathrm{Zn}_{(s)} & \rightarrow & \mathrm{Zn}^{2+} + 2 \ e^ & \text{(oxidation)}\\ \end{array} \end{align*}

When a halfreaction is reversed in direction, the sign of the \begin{align*}E^\circ\end{align*} value must be changed. Therefore, the two halfreactions, along with their \begin{align*}E^\circ\end{align*} values, become:


 \begin{align*} \begin{array}{rllr} \mathrm{Cu}^{2+} + 2 \ e^ & \rightarrow & \mathrm{Cu}_{(s)} & E^\circ = + 0.34 \ \text{volts}\\ \mathrm{Zn}_{(s)} & \rightarrow & \mathrm{Zn}^{2+} + 2 \ e^ & E^\circ = + 0.76 \ \text{volts}\end{array} \end{align*}

When the two halfreactions are added, we get both the balanced equation and the standard cell voltage.


 \begin{align*} \begin{array}{rllr} \mathrm{Cu}^{2+} + 2 \ e^ & \rightarrow & \mathrm{Cu}_{(s)} & E^\circ = + 0.34 \ \text{volts}\\ \mathrm{Zn}_{(s)} & \rightarrow & \mathrm{Zn}^{2+} + 2 \ e^ & E^\circ = + 0.76 \ \text{volts}\\ \hline \mathrm{Zn}_{(s)} + \mathrm{Cu}^{2+} & \rightarrow & \mathrm{Zn}^{2+} + \mathrm{Cu}_{(s)} & E^\circ_{\mathrm{NET}} = + 1.10 \ \text{volts} \end{array} \end{align*}

Example:
For the standard cell involving the dichromate half cell (\begin{align*}\mathrm{Cr}_2\mathrm{O}_7^{2} + 14 \ \mathrm{H}^+ + 6 \ e^ \rightarrow 2 \ \mathrm{Cr}^{3+} + 7 \ \mathrm{H}_2\mathrm{O}\end{align*}) whose \begin{align*}E^\circ\end{align*} is \begin{align*}+ 1.33 \ \mathrm{volts}\end{align*}, and the silver halfcell (\begin{align*}\mathrm{Ag}^+ + e^ \rightarrow \mathrm{Ag}_{(s)}\end{align*}) whose \begin{align*}E^\circ\end{align*} is \begin{align*}+ 0.80 \ \mathrm{volts}\end{align*}, determine what will be oxidized and reduced, name the oxidizing and reducing agents, find the \begin{align*}E{^{\circ}}_{\mathrm{NET}}\end{align*} for the cell, and balance the net equation.
Solution:
When we find the two halfreactions in the standard potential chart, the dichromate halfreaction is closer to the top of the chart, which means that it will run forward (reduction). The silver halfreaction, then, will be forced to reverse its direction (oxidation). We can write the two halfreactions in the direction they will run with the adjusted \begin{align*}E^\circ\end{align*} sign for the silver halfcell.


 \begin{align*} \begin{array}{rllr} \mathrm{Cr}_2\mathrm{O}_7^{2} + 14 \ \mathrm{H}^+ + 6 \ e^ & \rightarrow & 2 \ \mathrm{Cr}^{3+} + 7 \ \mathrm{H}_2\mathrm{O} & E^\circ = + 1.33 \ \text{volts}\\ \mathrm{Ag}_{(s)} & \rightarrow & \mathrm{Ag}^{+} + e^ & E^\circ =  0.80 \ \text{volts}\\ \end{array} \end{align*}

When we check the number of electrons on the two sides of the equation, we recognize that we must multiply the silver halfreaction by 6 in order for the electrons to cancel out. This brings up a question – when we multiply the halfreaction by 6, do we multiply the \begin{align*}E^\circ\end{align*} by 6? In the case of standard halfcells, in order for the halfcell to be standard, it must be at \begin{align*}25^\circ\mathrm{C}\end{align*} and the concentration of the ions must be \begin{align*}1.00 \ \mathrm{M}\end{align*}. Hence, when you multiply a halfreaction in order to balance an equation, you may be doubling an amount of solution, but the molarity must remain exactly \begin{align*}1.00 \ \mathrm{M}\end{align*}. Therefore, when you multiply a standard halfcell for balancing purposes, the \begin{align*}E^\circ\end{align*} remains exactly the same. The only instance where the \begin{align*}E^\circ\end{align*} of a standard halfcell changes is when the sign is changed to reverse the direction of the reaction. The halfreactions and net reaction for this problem become:


 \begin{align*} \begin{array}{rllr} \mathrm{Cr}_2\mathrm{O}_7^{2} + 14 \ \mathrm{H}^+ + 6 \ e^ & \rightarrow & 2 \ \mathrm{Cr}^{3+} + 7 \ \mathrm{H}_2\mathrm{O} & E^\circ = + 1.33 \ \text{volts}\\ 6 \ \mathrm{Ag}_{(s)} & \rightarrow & 6 \ \mathrm{Ag}^{+} + 6 \ e^ & E^\circ =  0.80 \ \text{volts}\\ \hline 6 \ \mathrm{Ag}_{(s)} + \mathrm{Cr}_2\mathrm{O}_7^{2} + 14 \ \mathrm{H}^+ & \rightarrow & 6 \ \mathrm{Ag}^{+} + 2 \ \mathrm{Cr}^{3+} + 7 \ \mathrm{H}_2\mathrm{O} & E^\circ_{\mathrm{NET}} = + 0.53 \ \text{volts}\\ \end{array} \end{align*}

The chromium in \begin{align*}\mathrm{Cr}_2\mathrm{O}_7^{2}\end{align*} is being reduced and is therefore the oxidizing agent. It is acceptable to just say the dichromate ion is the oxidizing agent rather than specifying the chromium in the dichromate ion.
The silver metal is being oxidized and is as a result the reducing agent.
This may be a convenient place to mention that standard cells last for only some small fraction of second when you turn them on. When cells begin to run, the ions on the reactant side of the equation are being used up, and the ions on the product side of the equation are being produced. When a standard cell begins to run, the molarities of the ions immediately change and are no longer exactly \begin{align*}1.00 \ \mathrm{M}\end{align*}. Consequently, the cell is no longer a standard cell.
Electrochemical Reactions and Metallic Corrosion
The corrosion of metals is usually an electrochemical reaction. The reaction is usually between a metal and the oxygen and water in the surroundings. Iron is a material that readily rusts (corrodes) in contact with oxygen and moisture. Since iron is an extremely common metal used for many purposes, such as building construction, doors, fencing materials, ships hulls, bolts, screws, and nails, considerable effort is spent trying to keep iron from rusting. There are several common ways of protecting iron from corrosion.
The most common way of protecting iron objects from corrosion is to paint them so that the oxygen and moisture cannot touch the metal. This system would include storing the iron objects in thick grease to keep possible reactants away. In certain cases, iron can be alloyed with other metals to make the product more resistant to corrosion. Stainless steel is an alloy of iron and small amounts of chromium. The addition of the small amount of chromium makes the steel less susceptible to corrosion. The addition of small amounts of other substances when alloying with iron also yields other positive properties.
Metallic iron may also be coated with a protective layer of a more reactive metal. Iron objects are often electroplated with chromium or zinc to provide the protective layer. Sometimes iron objects are dipped in molten zinc to produce a layer of more reactive metal. Covering a metal with a layer of zinc is called galvanizing and is a fairly common practice with fence wire, nails, buckets, and tubs. To protect ship hulls from corrosion, it is common to place a large mass of magnesium in electrical contact with the iron hull. The magnesium will corrode more easily than iron, so the reactants that contact the hull will react with the magnesium first, leaving the hull intact. This system is called cathodic protection.
Lesson Summary
 The redox reaction in an electrochemical cell is a spontaneous reaction.
 Ion flow between the chambers of an electrochemical cell keeps the chambers neutral.
 Standard potentials for halfcells are determined by measuring the potential of an electrochemical cell in which the tested cell is paired with the standard hydrogen halfcell.
 All standard potentials are measured at standard cell conditions of \begin{align*}25^\circ\mathrm{C}\end{align*} and \begin{align*}1.00 \ \mathrm{M}\end{align*} solutions of all ions.
 The voltage of a standard electrochemical cell is obtained by finding the difference between the reduction potentials of the two halfcells.
Further Reading / Supplemental Links
The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos. You are required to register before you can watch the videos but there is no charge. The website has one video that relates to this lesson called “The Busy Electron” that explains the principles of electrochemical cell design through batteries, sensors, and a solarpowered car.
The following provides text and animations about electrochemical cells and batteries.
Review Questions
 In a standard cell composed of the zinc and copper halfcells, will the current in the external circuit flow from \begin{align*}\mathrm{Zn}\end{align*} to \begin{align*}\mathrm{Cu}\end{align*} or from \begin{align*}\mathrm{Cu}\end{align*} to \begin{align*}\mathrm{Zn}\end{align*}?
 For a standard cell with the following balanced equation: \begin{align*}\mathrm{Sn}_{(s)} + 2 \ \mathrm{Ag}^+ \rightarrow \mathrm{Sn}^{2+} + 2 \ \mathrm{Ag}_{(s)}\end{align*},
 what is being oxidized?
 what is the reducing agent?
 what is the \begin{align*}E{^{\circ}}_{\mathrm{NET}}\end{align*} ?
 after the reaction has reached equilibrium, what will be its voltage?
 Balance the following equation using the halfreactions from the standard reduction potential chart: \begin{align*}\mathrm{Cr}_2\mathrm{O}_7^{2} + \mathrm{Fe}^{2+} + \mathrm{H}^+ \rightarrow \mathrm{Cr}^{3+} + \mathrm{Fe}^{3+} + \mathrm{H}_2\mathrm{O}\end{align*}.
All images, unless otherwise stated, are created by the CK12 Foundation and are under the Creative Commons license CCBYNCSA.
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