Electrochemistry (article) | Khan Academy (2024)

We encounter electrochemical cells in all facets of our everyday lives from the disposable AA batteries in our remote controls and the lithium-ion batteries in our iPhones to the nerve cells strewn throughout our bodies. There are two types of electrochemical cells: galvanic, also called Voltaic, and electrolytic. Galvanic cells derives its energy from spontaneous redox reactions, while electrolytic cells involve non-spontaneous reactions and thus require an external electron source like a DC battery or an AC power source. Both galvanic and electrolytic cells will consist of two electrodes (an anode and a cathode), which can be made of the same or different metals, and an electrolyte in which the two electrodes are immersed.

Galvanic cells traditionally are used as sources of DC electrical power. A simple galvanic cell may contain only one electrolyte separated by a semi-porous membrane, while a more complex version involves two separate half-cells connected by a salt bridge. The salt bridge contains an inert electrolyte like potassium sulfate whose ions will diffuse into the separate half-cells to balance the building charges at the electrodes. According to the mnemonic “Red Cat An Ox”, oxidation occurs at the anode and reduction occurs at the cathode. Since the reaction at the anode is the source of electrons for the current, the anode is the negative terminal for the galvanic cell.

Let’s look at an example of a galvanic cell like the classic AA alkaline battery, in which the equations for the two half-reactions are listed below:

$Mn{O}_{2(s)}+H_{2}O+e^{-}\to MnOO{H}_{(s)}+O{H}_{(aq)}^{-};E^o=+0.382\text{V}$‍$Zn(OH{)}_{2(s)}+2e^{-}\to Z{n}_{(s)}+2O{H}_{(aq)}^{-};E^o=-1.221\text{V}$‍

Here we are given two reduction potentials for the anode and cathode. Most data tables listing cell potentials will list reduction potentials. Since they are both reduction potentials, we need to decide which one needs to be flipped. Let’s keep in mind that when we flip the reduction potential into an oxidation potential, we also need to flip the sign. Since galvanic cells have a positive EMF, we are looking to flip the equation that when added to the other EMF will give us a positive value. By flipping the zinc half-reaction, we have the two EMF values as +0.382 V and +1.221 V. To find the cell EMF, we simply add them together to give us an approximate value of 1.5 V, which we already know to be the EMF of an alkaline AA battery.

Since voltage is an intensive property, which is one that does not depend on the system size or the amount of material in the system, we do not have to multiply the EMF by any stoichiometric coefficient to cancel out the electrons in calculating EMF.

Suppose we are asked to put the amount of energy into more thermodynamic terms. Let’s use the following equation to connect E${}^{\text{o}}$‍ to ∆G${}^{\text{o}}$‍ :

where n is the moles of electrons transferred, E${}^{\text{o}}$‍ is the standard state EMF, and F is Faraday’s constant, 96,485 C/mol e${}^{\text{–}}$‍, but we will approximate as 10${}^{\text{5}}$‍ C/mol e${}^{\text{–}}$‍. One way to remember this equation is to compare it to the formula for electric potential energy, U = qV. U can be correlated with ∆G${}^{\text{o}}$‍, both expressed in joules, and E${}^{\text{o}}$‍ with V, both expressed in volts. nF cancels out to be coulombs, just as q is.

We can start by plugging in values for an alkaline battery into our equation for the standard Gibbs free energy:

$∆G^o=-nF{E}^o$‍
$∆G^o=-(2\text{mol}{e}^{-})\left({10}^5{\displaystyle \frac{C}{\text{mol}{e}^{-}}}\right)(1.5\text{V})$‍
$∆G^o=-300,000\text{J or}-300\text{kJ}$‍

Gibbs free energy is normally expressed in kilojoules rather than joules. The sign gives us an indication of the direction that the reaction has to shift to reach equilibrium. This means that a system under standard conditions would have to shift to the right, converting some reactants into products before reaching equilibrium. The magnitude gives us an indication of how far the standard state is from equilibrium. The larger the value, the further the reaction has to shift from standard state conditions to reach equilibrium. A negative value for ∆G${}^{\text{o}}$‍ also indicates the spontaneity of the reaction.

Suppose we are asked to calculate the equilibrium constant K at standard conditions to get an indication of how favorable this reaction is.

$∆G^o=-{\text{RT ln K}}_{eq}$‍
$∆G^o=-\left(8.31{\displaystyle \frac{J}{mol\ast K}}\right)(298\text{K})(2.303)log\text{K}$‍

Let’s approximate the gas constant R with the value of 8.31 J/(mol∙K) as 8 and the standard temperature of 298 K as 300, which when multiplied together calculates to a value of 2400. Since we rounded down the original value of R earlier, the actual calculated value should be closer to 2500, which we will use instead. We need to multiply R by the conversion factor for lnx to logx, which is 2.303. We would not be expected to know this value and would most likely be provided the conversion factor, which we will round down to 2.3. In order to calculate (2500 x 2.3), use the values of 2500 x 2 and 2500 x 3 as the lower and upper boundaries, respectively 5000 and 7500. 2.3 is closer to 2 than 3, so let’s find a value closer to 5000 than 7500, which would be 6000.

We can plug in the value of ∆G${}^{\text{o}}$‍ on the left side of the equation. Even though ∆G${}^{\text{o}}$‍ is normally expressed as kJ/mol, R is expressed as J/mol∙K, so we can convert R or ∆G${}^{\text{o}}$‍ to match units. Let’s plug in 300,000 J for ∆G${}^{\text{o}}$‍ to match R. Divide 300,000 by 6,000 to obtain a value of 50. To get rid of the logarithm, we raise both sides as exponents with a base of 10. Our calculated equilibrium constant is approximately 10${}^{\text{50}}$‍, while the actual value is 10${}^{\text{52}}$‍. In matching our answer to an answer choice, we are not looking an exact match. For example, answer choices that we may encounter are the following: 10${}^{\text{52}}$‍, 10${}^{\text{-52}}$‍, 10${}^{\text{42}}$‍, and 10${}^{\text{62}}$‍.

The large K value indicates that the reaction is very favorable towards the products and will go entirely to completion. In the case of the batteries, the reaction will run until it reaches equilibrium, i.e. ∆G${}^{\text{o}}$‍ = 0.

Let’s consider a concentration cell, which is a specific form of a galvanic cell with two equivalent half-cells of the same material differing only in concentration. We can find concentration cells in the concentration gradients in our nerve cells, the Na${}^{\text{+}}$‍/K${}^{\text{+}}$‍ or Ca${}^{\text{2+}}$‍ ion pumps in our cell membranes, and ATP synthase used in energy production.

A concentration cell produces a small voltage as it attempts to reach equilibrium. We can calculate the potential developed by a concentration cell using the Nernst Equation, which is as follows:

where E is cell potential of interest, E${}^{\text{o}}$‍ is the standard cell potential, R is the gas constant, T is the absolute temperature, n is the moles of electrons transferred in the reaction, F is Faraday’s constant, and Q is the reaction quotient.

We can take note how the Nernst equation fits the rubric for any equation with such a wide-ranging set of values that requires that it be expressed in a logarithmic scale: calculated value = standard or reference value ± logarithmic term. Let’s compare this to other equations with a logarithmic term, like the Henderson-Hasselbalch equation and, of course, the thermodynamics equation from which the Nernst equation is derived:

$pH=p{K}_a+log{\displaystyle \frac{[A]}{[HA]}}$‍
$∆G=∆G^o+\text{RT lnQ}$‍

We can take notice how the equation matches the algebraic formula for a linear graph: y = mx + b, where y is E, m is -RT/nF, x is lnQ, and b is E${}^{\text{o}}$‍.

For concentration cells, the Nernst equation should be simplified to the following:

since (RT/F)(ln conversion factor) is equivalent to 0.0592 V, which we will round up to 0.06 V. As for the logarithmic term, [ion]/[ion] represents the reaction quotient Q. When Q = 1, the cell reaches equilibrium and is considered “dead” since the two concentrations are equal. In addition, log(1) = 0 so that E = 0, which confirms that the cell would be “dead”. When [ion]${}_{low}$‍ is in the numerator, the fraction will be less than 1, while when [ion]${}_{high}$‍ is in the numerator, the fraction will be greater than 1. We want the configuration in which the fraction is less than 1 since Q < K${}_{eq}$‍, the reaction will move towards equilibrium, and here, K${}_{eq}$‍ would be 1.

Suppose we have a concentration cell with 0.005 M Cu${}^{\text{2+}}$‍ in one cell and 0.10 M Cu${}^{\text{2+}}$‍ in another cell, we are asked to calculate the EMF of the concentration cell:

For any concentration cell, the standard state EMF (E${}^{\text{o}}$‍) is 0 since the two half-cells have the same half-reactions. Next, we substitute the concentration value of 0.005 M into the numerator and 0.1 M into the denominator since the smaller value goes in the numerator. We should also keep in mind that Cu is a 2+ ion, so the value of n, which is the moles of electrons, is 2. During our calculation, we can also use the following logarithmic rule: log(ab) = log a + log b. Ultimately, we get the value of 0.0392 V.

One way of approaching logarithmic values without calculators is to use the below approximation. We can use the sequential odd numbers 3, 5, 7, and 9 to help us approximate the logarithmic values between 1 and 10, as in 3-5, 5-7, and 7-9 or log(3) - 0.5, log(5) - 0.7, and log(7) - 0.9.

If we are given log(0.073) or log(55), we should approach the calculation as follows:

Let’s take a look at the final type of cell, the electrolytic cell. An electrolytic cell is an electrochemical cell in which the energy from an external power source is used to drive a normally non-spontaneous reaction, i.e. apply a reverse voltage to a voltaic cell. We encounter electrolytic cells during the charging phase of any type of rechargeable battery from the lead-acid battery in automobiles to the lithium-ion battery in smartphones.

In comparison to the galvanic cell, the electrodes of an electrolytic cell can be placed in a single compartment containing the molten or aqueous electrolyte. In addition, since the external battery source is what drives the electrons through the circuit, the electrodes will match the positive and negative terminal of the battery. While the anode remains the site of oxidation, it becomes the positive terminal, and the cathode becomes the negative terminal.

Another use for an electrolytic cell is for the decomposition of compounds, i.e. water, sodium chloride, into simpler compounds. Industrial processes take advantage of this in the production of chlorine or sodium hydroxide. Since electrolytic cells can be conducted in molten or aqueous electrolyte, depending on the cation and anion, the products from using a molten electrolyte may be different from the products from using an aqueous electrolyte.

In aqueous solution, at the cathode or negative electrode, if a metal is more reactive than hydrogen, then the reduced metal reacts with water to produce hydrogen gas. We can look to the activity series or a table of reduction potential to determine that the following cations are harder to reduce than water: Group IA (Li${}^{\text{+}}$‍, Rb${}^{\text{+}}$‍, K${}^{\text{+}}$‍, Cs${}^{\text{+}}$‍, Na${}^{\text{+}}$‍) and Group IIA (Ba${}^{\text{2+}}$‍, Sr${}^{\text{2+}}$‍, Ca${}^{\text{2+}}$‍, Mg${}^{\text{2+}}$‍) cations. However, if a metal is less reactive than hydrogen, then the reduced metal does not react with water, i.e. copper, platinum, etc.

When looking at the electrolysis of sodium chloride in aqueous solution, there are additional reactions we must consider in calculating EMF. At the anode, the following reactions could occur:

$2C{l}^{–}\to C{l}_2(g)+2e^{–}\text{;}{E}^o=-1.36\text{V}$‍
$2H_{2}O\to O_2(g)+4H^{+}+4e^{–}\text{;}{E}^o=-1.23\text{V}$‍

While at the cathode, the following reactions could occur:

$N{a}^{+}+e^{–}\to Na(s)\text{;}{E}^o=-2.7\text{V}$‍
$2H_{2}O+2e^{–}\to H_2(g)+2O{H}^{–}\text{;}{E}^o=-0.83\text{V}$‍

At the anode, the oxidation of hydroxide is kinetically hindered because reactions involving oxygen gas are very slow. At the cathode, the reduction of water is more energetically favorable since it has a more positive reduction potential. NaOH is produced instead of sodium metal.

Suppose we are asked to calculate the EMF of the above electrolytic cell. We were given that the cell creates chlorine and sodium hydroxide, so we must pick these two reactions:

$2C{l}^{–}\to C{l}_2(g)+2e^{–}\text{;}{E}^o=-1.36\text{V}$‍
$2H_{2}O+2e^{–}\to H_2(g)+2O{H}^{–}\text{;}{E}^o=-0.83\text{V}$‍

Normally we have to flip one of the equations into an oxidation potential, but the reaction is already written as an oxidation potential. We will most likely be given reduction potentials in a chart or table, but sometimes oxidation potentials are given. Adding the two EMF values together, the total E${}^{\text{o}}$‍ for this electrolytic cell is -2.19 V, so it will require 2.19 V to drive this reaction.

In order to produce sodium or aluminum, we must use the electrolyte in molten state. For an electrolyte as molten sodium chloride, the following half-reactions take place:

$N{a}^{+}+e^{–}\to Na(s)\text{;}{E}^o=-2.7\text{V}$‍
$2C{l}^{–}\to C{l}_2(g)+2e^{–}\text{;}{E}^o=-1.36\text{V}$‍

In comparison to -2.19 V, the EMF for this cell is -4.0 V.

Another common use of electrolysis is in electroplating, which always occurs at the cathode. We can calculate the amount of metal plated on the cathode from the charge or current that passes through the cell, and in fact they are proportional in the following relationship:

where I is the current expressed in amperes or coulombs/second, n is the moles of electrons and F is Faraday’s constant, which has the value of 96,485 C/mol e${}^{\text{–}}$‍. We should always approximate Faraday’s constant as 10${}^{\text{5}}$‍ for our purposes in calculations.

Suppose we are asked to figure out the amount of platinum metal in grams that is deposited when 5.0 amps of current is passed through the electrolyte for 1 hour. Applying the equation, we plug in the following values:

$(5.0\text{A})(60\text{min})\left({\displaystyle \frac{60\text{sec}}{1\text{min}}}\right)=n(96,485{\displaystyle \frac{C}{\text{mol}{e}^{-}}})$‍
$(5)(3600)=n(100000)$‍
$(18000)=n(100000)$‍
$n=0.18\text{mol}e$‍
$(0.18\text{mol}e)\left({\displaystyle \frac{1\text{mol}Pt}{2\text{mol e}}}\right)=0.09\text{mol}Pt$‍
$(0.09\text{mol}Pt)\left(195{\displaystyle \frac{gPt}{\text{mol}Pt}}\right)\approx (0.09\text{mol}Pt)\left(200{\displaystyle \frac{gPt}{\text{mol}Pt}}\right)=18gPt$‍

As expected, we approximated Faraday’s constant as 100,000. We also have to convert 1 hour into seconds, which in total is 3600 seconds. The following math is straightforward and does not require much rounding. The only tricky part is remembering that n represents the mole of electrons. Since platinum forms a 2+ ion, then there are two moles of electrons consumed. To obtain the moles of platinum produced, we need to divide by 2. At the end, we can approximate 195 as 200 to obtain our final value of 18 grams.

We may expect to see 18.2 g, 36.4 g, 1.82 g and 3.64 g as possible answer choices to account for losing a power of 10 in our calculations or forgetting to divide by 2 to factor in that n represents the moles of electrons not the moles of the metal. Since there was little rounding, our calculated answer should pretty closely match one of the answer choices.

Here is a table that highlights the important differences between electrolytic and galvanic cells that we need to know for the MCAT exam.