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What do you think of when you think of a battery? Maybe a remote control, or your phone, or even a car? Well, what about a 2000-year-old clay pot? Believe it or not, around the year 150 BC in Iraq and some other areas, civilizations knew how to harness electricity into a battery. Somehow lost to the sands of time, this technology wasn't again repeated until 1799! So, how does a battery even operate, and how did these ancient peoples do it? Whether they knew it or not, they used a little concept called standard potential. This concept—and more—will be covered here, so buckle up and prepare to be electrified!
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Jetzt kostenlos anmeldenWhat do you think of when you think of a battery? Maybe a remote control, or your phone, or even a car? Well, what about a 2000-year-old clay pot? Believe it or not, around the year 150 BC in Iraq and some other areas, civilizations knew how to harness electricity into a battery. Somehow lost to the sands of time, this technology wasn't again repeated until 1799! So, how does a battery even operate, and how did these ancient peoples do it? Whether they knew it or not, they used a little concept called standard potential. This concept—and more—will be covered here, so buckle up and prepare to be electrified!
With the discovery of electricity, in 1799, Alessandro Volta created the Voltaic Pile, the first modern battery. Volta created the first modern battery, which is now often called a Galvanic (or Voltaic) cell. The Galvanic cell uses an electrical potential difference to create electricity.
Standard potential refers to the chemistry which takes place in a Galvanic (Voltaic) cell, under standard conditions.1 These cells exclusively house Thermodynamically Favored reactions. We know this because of calculations determining Cell Potential, which is related to Gibbs Free Energy, and the Equilibrium Constant or Reaction Quotient. Thermodynamically unfavourable reactions are also possible, but they are only contained in electrolytic cells, which is provided through the study of Electrolysis.
In a Galvanic cell, we produce a Redox reaction that has two species— an oxidant, and a reductant. Recall that an oxidant will oxidize something else (meaning that it is reduced— or takes electrons). Conversely, a reductant is something that reduces something else (meaning that it is oxidized— or gives away electrons). So, in a Redox reaction, there are two reactions taking place: one reaction that is being oxidized, and one reaction that is being reduced.
In redox chemistry, the standard electrode half-cell potentials are the cell potential values that determine whether each half-reaction will be oxidized or reduced. Some molecules like electrons more than others, so some molecules want to take electrons, and some wish to give them away. This phenomenon is something that can be quantified and has been by many chemists over the past century.
What this boils down to, is that each reaction has its own standard cell potential value (in Volts), which measures how likely it is to be reduced. But how are the reduction potentials of each reaction determined? Every reaction is compared to a standard reaction, which is referred to as the standard electrode. This is known as the Standard Hydrogen Electrode (S.H.E.).
The standard hydrogen electrode consists of hydrogen reduction in the presence of a platinum electrode. Hydrogen gas is bubbled through a solution of strong acid, at pH = 0. The hydrogen gas reaches an equilibrium with the hydrogen ions and pressure remains at 1 bar. Another half-cell is combined with the hydrogen electrode with a voltmeter. The reaction that occurs is either the reduction of protons (hydrogen ions) to form hydrogen gas, or the oxidation of hydrogen gas to form protons.
$$ 2 H^+ ( aq, ~ 1 ~ mol ~ L ^ {-1} ) + 2 e^- \rightleftharpoons H_2 ( g, ~ 1 ~ bar ) $$
The standard hydrogen electrode has been universally declared to reside at a potential of 0 V. Therefore:
$$ E^ {\circ} _ { H^+ / H_2 } = 0 ~ V $$
Remember that the standard reduction potential of a half-cell does not represent its thermodynamic favourability. Not alone, at least. Only when combined with another reaction can you determine whether a redox reaction is thermodynamically favorable or not.
The standard hydrogen electrode is considered standard because the acid Concentration is at 1 mol L-1, and the pressure is 1 bar. However, at different pH, the electrode has a different reduction potential. This can be determined by using a variation of the Nernst equation.
$$ E _{cell} = E ^ {\circ} _ {cell} - \frac {0.059} {n} \log {Q} $$
Recall that the reduction reaction for hydrogen is:
$$ 2 H^+ (aq) + 2 e^- \rightleftharpoons H_2 (g) \qquad E ^ {\circ} _ {cell} = 0 ~ V $$
Since pH = -log[H+], this can be applied in the equation.
\begin{align}&E _{cell} = E ^ {\circ} _ {cell} - \frac {0.059} {n} \log { \frac { P (H_2) } { [H^+]^2 } } \\&E _{cell} = E ^ {\circ} _ {cell} - \frac {0.059} {n} \log { \frac { 1 } { [H^+]^2 } } \\&E _{cell} = E ^ {\circ} _ {cell} - \frac {0.059} {n} \log { [H^+]^{-2 } } \\&E _{cell} = 0 - \frac {0.059} {2} ( -2 \log { [H^+] } ) \\&E _{cell} = -0.059 ~ pH\end{align}
At pH = 0, Ecell = E°cell = 0 V. However, at pH = 7, Ecell = -0.41 V. This means that acids are more oxidizing when they are more concentrated.
By using hydrogen's half-cell reaction, chemists have come up with tables that display the relative reduction rates of many reactions. To get an idea of what one looks like, an example of a reduction table is provided.
A reduction standard potential table will list half-cell potentials in ascending order. The highest values (most negative) represent the stronger reducing agents. At the bottom of the table, the more positive values will represent the stronger oxidizing agents. This means that the more negative values are more likely to be the anode, and the more positive values more likely to be the cathode. Just remember that these values are only relative to the other half cell. Meaning, if you have two strong oxidants reacting, the stronger one will inevitably be reduced.
| Oxidized form + electron(s) | $$\bf \rightarrow $$ | Reduced form | $$ \bf E^ {\circ} _ {cell} $$ |
| $$ Li^ + (aq) + e^- $$ | $$\bf \rightarrow $$ | $$ Li (s) $$ | $$ E^{\circ}_{cell} = -3.04~V $$ |
| $$ K^+ (aq) + e^- $$ | $$\bf \rightarrow $$ | $$ K (s) $$ | $$ E^{\circ}_{cell} = -2.93~V $$ |
| $$ Ca ^{2+} (aq) + 2 e^- $$ | $$\bf \rightarrow $$ | $$ Ca (s) $$ | $$ E^{\circ}_{cell} = -2.87~V $$ |
| $$ Al ^ {3+} (aq) + 3 e^- $$ | $$\bf \rightarrow $$ | $$ Al (s) $$ | $$ E^{\circ}_{cell} = -1.66~V $$ |
| $$ Zn ^ {2+} (aq) + 2e^- $$ | $$\bf \rightarrow $$ | $$ Zn (s) $$ | $$ E^{\circ}_{cell} = -0.76~V $$ |
| $$ Fe ^{2+} (aq) + 2e^- $$ | $$\bf \rightarrow $$ | $$ Fe (s) $$ | $$ E^{\circ}_{cell} = -0.44~V $$ |
| $$ Fe ^ {3+} (aq) + 3e^- $$ | $$\bf \rightarrow $$ | $$ Fe (s) $$ | $$ E^{\circ}_{cell} = -0.04~V $$ |
| $$ 2 H^+ + 2e ^- $$ | $$\bf \rightarrow $$ | $$ H_2 (g) $$ | $$ E^{\circ}_{cell} = 0~V $$ |
| $$ Cu ^ {2+} (aq) + e^- $$ | $$\bf \rightarrow $$ | $$ Cu ^ + (aq) $$ | $$ E^{\circ}_{cell} = +0.15~V $$ |
| $$ Cu ^ {2+} (aq) + 2 e^- $$ | $$\bf \rightarrow $$ | $$ Cu (s) $$ | $$ E^{\circ}_{cell} = +0.34~V $$ |
| $$ Fe ^ {3+} (aq) + e^- $$ | $$\bf \rightarrow $$ | $$ Fe^ {2+} $$ | $$ E^{\circ}_{cell} = +0.77~V $$ |
| $$ Ag^+ (aq) + e^- $$ | $$\bf \rightarrow $$ | $$ Ag(s) $$ | $$ E^{\circ}_{cell} = +0.80~V $$ |
| $$ O_2 (g) + 4 H^+ (aq) + 4e^- $$ | $$\bf \rightarrow $$ | $$ H_2O (l) $$ | $$ E^{\circ}_{cell} = +1.23~V $$ |
| $$ Co ^ {3+} (aq) + e^- $$ | $$\bf \rightarrow $$ | $$ Co ^ {2+} (aq) $$ | $$ E^{\circ}_{cell} = +1.92~V $$ |
| $$ F_2 (aq) + 2e^- $$ | $$\bf \rightarrow $$ | $$ 2 F^- (aq) $$ | $$ E^{\circ}_{cell} = +2.87~V $$ |
This table is a shortened version of an actual reduction table. By using one of these, you can determine the thermodynamic viability of a reaction when combining any two half-reactions. However, it is important to remember that this will show the spontaneity of a reaction, but not the speed of reaction.
It is a general rule of thumb that a reaction will run quickly if it has a nonstandard cell potential, Ecell > 0.60 V. In general, the higher the gap in potential (called an "overpotential"), the faster the reaction will run.
Now that we have a reduction table, we can start combining these reactions and see what is Thermodynamically Favored. The standard potential equation has been introduced before; however, it will be introduced again as a quick reminder. When combining the standard cell potentials of two half-cells, we acquire the overall cell potential. This is done as follows:
$$ E ^ {\circ} _ {cell} = E ^ {\circ} _ {cathode} - E ^ {\circ} _ {anode} $$
The values that are written in the reduction table are the tendency of that reaction to be reduced. These are the values that are plugged into the formula. The species which will occupy the cathode in a Galvanic cell is the one that is being reduced. Conversely, the anode will be oxidized. Once you have determined which species will be which, you can combine them into a full reaction. Just remember that the oxidation reaction will need to be flipped so that it represents an oxidation, and not a reduction (as it is listed in the table).
For example, let's try reacting Cu(II) and Zn(II) in a Galvanic cell. First, let's write down each of their half-reactions.
\begin{align}Zn ^ {2+} (aq) + 2e ^- &\rightarrow Zn (s) \qquad &E^ {\circ} _ {cell} &= -0.76 ~ V \\Cu ^ {2+} (aq) + 2e ^- &\rightarrow Cu (s) \qquad &E^ {\circ} _ {cell} &= +0.34 ~ V\end{align}Zinc has more negative cell potential, so it will be the stronger reducing agent. This means that zinc will be the anode, and copper will be the cathode. Since zinc is being oxidized, we will need to flip around the reaction to combine them and cancel our like terms.
\begin{align}Zn (s) &\rightarrow Zn ^ {2+} (aq) + 2e ^- \\Cu ^ {2+} (aq) + 2e ^- &\rightarrow Cu (s) \\Zn (s) + Cu ^ {2+} (aq) + 2e ^- &\rightarrow Zn ^ {2+} (aq) + Cu (s) + 2e ^- \\Zn (s) + Cu ^ {2+} (aq) &\rightarrow Zn ^ {2+} (aq) + Cu (s)\end{align}
Now that we have our full reaction, we can test its thermodynamic favourability using our equation of standard potential.
\begin{align}&E ^ {\circ} _ {cell} = E ^ {\circ} _ {cathode} - E ^ {\circ} _ {anode} \\&E ^ {\circ} _ {cell} = E ^ {\circ} _ { Cu ^{2+} / Cu } - E ^ {\circ} _ { Zn ^{2+} / Zn } \\&E ^ {\circ} _ {cell} = 0.34 ~ V - ( -0.76 ~ V ) \\&E ^ {\circ} _ {cell} = + 1.10 ~ V\end{align}
So, the two half-cells we selected will indeed react and will do so rather quickly.
Recognizing whether a reaction is a redox reaction, and then determining which species is being reduced, and which is being oxidized, takes some practice. This is a lot easier thanks to a reduction potential table. However, it is still necessary to take opportunities to practice deconstructing Redox Reactions whenever possible. Follow along with this example and judge to see whether you could do this process on your own.
Let's look at the reaction which represents the formation of water:
$$ 4 H_2 (g) + O_2 (g) \rightarrow 2 H_2O (l) \qquad E^ {\circ} _ {cell} = +1.23 ~ V $$
So, by looking at this, it appears that the formation of water is a thermodynamically favorable reaction. But, to fully understand it, let's break down the reaction into its two half-reactions.
\begin{align}2 H^+ + 2 e^- &\rightarrow H_2 (g) \qquad &E^ {\circ} _ {cell} &= 0 ~ V \\O_2 (g) + 4 H^+ + 4 e^- &\rightarrow 2 H_2O (l) \qquad &E^ {\circ} _ {cell} &= +1.23 ~ V\end{align}
Now that it's deconstructed, let's put this into our standard cell potential equation just to verify. Recall that the more positive cell potential will represent the stronger oxidizing agent. This means that oxygen will be the cathode, and hydrogen will be the anode.
\begin{align}&E ^ {\circ} _ {cell} = E ^ {\circ} _ {cathode} - E ^ {\circ} _ {anode} \\&E ^ {\circ} _ {cell} = E ^ {\circ} _ { O_2, H^+ / H_2O } - E ^ {\circ} _ { H^+ / H_2 } \\&E ^ {\circ} _ {cell} = +1.23 ~ V - 0.00~ V \\&E ^ {\circ} _ {cell} = + 1.23 ~ V\end{align}
Forming water from hydrogen and oxygen is a rapid and facile reaction. The opposite reaction, however, is quite difficult to achieve. This is referred to as water splitting, and is currently a very relevant topic of research. You may wonder why chemists are interested in splitting water, but the answer lies in the products. Which do you think is more important to produce: hydrogen or oxygen? Contrary to what you may think, the desired product in the splitting of water is hydrogen gas. This is because hydrogen offers a potential green fuel source for the future.
Galvanic and Electrolytic Cells are usually prepared with an aqueous media. Aqueous electrolyte solutions are so common because they are very effective in carrying electrical charge. Ions easily dissolve in water, which is what makes them a great conductors. So, by adding an electrolyte solution to water, electrons easily flow from the electrolyte to the electrode or vice versa.
Determining the standard potential of a solution is also almost always done in aqueous solution. Remember how we discussed the standard hydrogen electrode and its pH? Well, you may not have realized it at the time, but that was also in water. Most electrochemical cells, and thus most standard potentials, are done in an aqueous media. There are only a few situations in which aqueous solutions are not used.
Determining the standard potential in nonaqueous solutions is sometimes necessary. When one of the half-reactions will favorably react with water, it will disrupt the standard measurements. Li+ is a great example of this because it likes to react with water to form LiOH (s) and H2 (g).
Lithium ions also like to "chelate" with water, which means that they prefer to stay in close proximity to the water molecules. Both of these result in less redox chemistry with the other half-reaction in the cell. As a result, lithium reactions are not run in water, but are instead run in nonaqueous electrolyte solutions.
You've likely heard of lithium-ion batteries. They have become the go-to rechargeable battery in phones, laptops, and now electric cars. They have many advantages to them, but they must be prepared in nonaqueous solutions. This isn't necessarily a problem, though. Running an Electrochemical Cell in nonaqueous solution does some advantages. This allows the cell to be run at very high cell potentials, > 3 V.2 This would normally not be possible in aqueous solutions because this overpotential could oxidize water, instead of the electrode. Fortunately, standard potentials can be performed in nonaqueous solutions, providing a route to more extreme cell potentials.
Standard cell potential of a redox reaction is calculated using the standard reduction values of each half cell reaction. These are found in a reduction potential table, which can then be put into the following equation:
E°cell = E°cathode - E°anode
The standard reduction potential is the tendency of a half reaction to be reduced. Each reaction has its own standard reduction potential, E°cell, in Volts (V). These values are determined by reacting the half reaction with a standard hydrogen electrode.
The potential of the standard hydrogen electrode (S.H.E.) is set to 0 to form a universal standard. The potential is 0 at pH = 0 and changes with varying pH. To calculate the S.H.E. potential at a different pH, the Nernst equation is used and [H+] is used in Q.
Ecell = E°cell - (0.059 / n)* logQ
To determine standard values for all other reactions, the standard hydrogen electrode (S.H.E.) is set to 0 V. This maintains proportionality among all half reactions, since each reaction will either oxidize or reduce the S.H.E.
Every half cell reaction will have its own standard electrode potential, which is the likelihood of this reaction being reduced. These values are compared to the standard hydrogen electrode (S.H.E.), which is set to 0 V.
In a Galvanic cell, zinc is reacted with S.H.E. The reaction produces bubbling of H2 (g), and a potential of 0.76 V.
E°cell = E°cathode - E°anode
E°cell = E°H+/H2 - E°Zn2+/Zn
E°cell = 0 V - 0.76 V
E°cell = 0.76 V for the half reaction of Zn2+ --> Zn(s)
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SIGNUP SIGNUPFlashcards in Standard Potential13
Start learningHow does a Galvanic (Voltaic) cell operate?
It uses a potential difference between the cathode and anode.
What is standard potential chemistry?
Electrohemical reactions which operate under standard conditions.
How does standard potential relate to Gibbs free energy?
They both represent the thermodynamic favorability of a reaction.
An oxidant will _____ another species, which means it _____ electrons.
oxidize, takes.
An reductant will _____ another species, which means it _____ electrons.
reduce, gives.
How is the standard potential of an electrode determined?
The reaction is reacted with the Standard Hydrogen Electrode in a Galvanic cell.
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