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By the end of this section, you will be able to:
  • Determine standard cell potentials for oxidation-reduction reactions
  • Use standard reduction potentials to determine the better oxidizing or reducing agent from among several possible choices

The cell potential in [link] (+0.46 V) results from the difference in the electrical potentials for each electrode. While it is impossible to determine the electrical potential of a single electrode, we can assign an electrode the value of zero and then use it as a reference. The electrode chosen as the zero is shown in [link] and is called the standard hydrogen electrode (SHE)    . The SHE consists of 1 atm of hydrogen gas bubbled through a 1 M HCl solution, usually at room temperature. Platinum, which is chemically inert, is used as the electrode. The reduction half-reaction chosen as the reference is

2H + ( a q , 1 M ) + 2 e H 2 ( g , 1 atm ) E ° = 0 V

E ° is the standard reduction potential. The superscript “°” on the E denotes standard conditions (1 bar or 1 atm for gases, 1 M for solutes). The voltage is defined as zero for all temperatures.

The figure shows a beaker just over half full of a blue liquid. A glass tube is partially submerged in the liquid. Bubbles, which are labeled “H subscript 2 ( g )” are rising from the dark grey square, labeled “P t electrode” at the bottom of the tube. A curved arrow points up to the right, indicating the direction of the bubbles. A black wire which is labeled “P t wire” extends from the dark grey square up the interior of the tube through a small port at the top. A second small port extends out the top of the tube to the left. An arrow points to the port opening from the left. The base of this arrow is labeled “H subscript 2 ( g ) at 1 a t m.” A light grey arrow points to a diagram in a circle at the right that illustrates the surface of the P t electrode in a magnified view. P t atoms are illustrated as a uniform cluster of grey spheres which are labeled “P t electrode atoms.” On the grey atom surface, the label “e superscript negative” is shown 4 times in a nearly even vertical distribution to show electrons on the P t surface. A curved arrow extends from a white sphere labeled “H superscript plus” at the right of the P t atoms to the uppermost electron shown. Just below, a straight arrow extends from the P t surface to the right to a pair of linked white spheres which are labeled “H subscript 2.” A curved arrow extends from a second white sphere labeled “H superscript plus” at the right of the P t atoms to the second electron shown. A curved arrow extends from the third electron on the P t surface to the right to a white sphere labeled “H superscript plus.” Just below, an arrow points left from a pair of linked white spheres which are labeled “H subscript 2” to the P t surface. A curved arrow extends from the fourth electron on the P t surface to the right to a white sphere labeled “H superscript plus.” Beneath this atomic view is the label “Half-reaction at P t surface: 2 H superscript plus ( a q, 1 M ) plus 2 e superscript negative right pointing arrow H subscript 2 ( g, 1 a t m ).”
Hydrogen gas at 1 atm is bubbled through 1 M HCl solution. Platinum, which is inert to the action of the 1 M HCl, is used as the electrode. Electrons on the surface of the electrode combine with H + in solution to produce hydrogen gas.

A galvanic cell consisting of a SHE and Cu 2+ /Cu half-cell can be used to determine the standard reduction potential for Cu 2+ ( [link] ). In cell notation, the reaction is

Pt ( s ) H 2 ( g , 1 atm ) H + ( a q , 1 M ) Cu 2+ ( a q , 1 M ) Cu ( s )

Electrons flow from the anode to the cathode. The reactions, which are reversible, are

Anode (oxidation): H 2 ( g ) 2H + ( a q ) + 2e Cathode (reduction): Cu 2+ ( a q ) + 2e Cu ( s ) ¯ Overall: Cu 2+ ( a q ) + H 2 ( g ) 2H + ( a q ) + Cu ( s )

The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is necessary because oxidation is the reverse of reduction.

E cell ° = E cathode ° E anode °
+0.34 V = E Cu 2+ /Cu ° E H + /H 2 ° = E Cu 2+ /Cu ° 0 = E Cu 2+ /Cu °
This figure contains a diagram of an electrochemical cell. Two beakers are shown. Each is just over half full. The beaker on the left contains a clear, colorless solution and is labeled below as “1 M H superscript plus.” The beaker on the right contains a blue solution and is labeled below as “1 M C u superscript 2 plus.” A glass tube in the shape of an inverted U connects the two beakers at the center of the diagram. The tube contents are colorless. The ends of the tubes are beneath the surface of the solutions in the beakers and a small grey plug is present at each end of the tube. The beaker on the left has a glass tube partially submersed in the liquid. Bubbles are rising from the grey square, labeled “Standard hydrogen electrode” at the bottom of the tube. A curved arrow points up to the right, indicating the direction of the bubbles. A black wire extends from the grey square up the interior of the tube through a small port at the top to a rectangle with a digital readout of “positive 0.337 V” which is labeled “Voltmeter.” A second small port extends out the top of the tube to the left. An arrow points to the port opening from the left. The base of this arrow is labeled “H subscript 2 ( g ).” The beaker on the right has an orange-brown strip that is labeled “C u strip” at the top. A wire extends from the top of this strip to the voltmeter. An arrow points toward the voltmeter from the left which is labeled “e superscript negative flow.” Similarly, an arrow points away from the voltmeter to the right. A curved arrow extends from the standard hydrogen electrode in the beaker on the left into the surrounding solution. The tip of this arrow is labeled “H plus.” An arrow points downward from the label “e superscript negative” on the C u strip in the beaker on the right. A second curved arrow extends from another “e superscript negative” label into the solution below toward the label “C u superscript 2 plus” in the solution. A third “e superscript negative” label positioned at the lower left edge of the C u strip has a curved arrow extending from it to the “C u superscript 2 plus” label in the solution. A curved arrow extends from this “C u superscript 2 plus” label to a “C u” label at the lower edge of the C u strip.
A galvanic cell can be used to determine the standard reduction potential of Cu 2+ .

Using the SHE as a reference, other standard reduction potentials can be determined. Consider the cell shown in [link] , where

Pt ( s ) H 2 ( g , 1 atm ) H + ( a q , 1 M ) Ag + ( a q , 1 M ) Ag ( s )

Electrons flow from left to right, and the reactions are

anode (oxidation): H 2 ( g ) 2H + ( a q ) + 2e cathode (reduction): 2 Ag + ( a q ) + 2e 2Ag ( s ) ¯ overall: 2 Ag + ( a q ) + H 2 ( g ) 2H + ( a q ) + 2Ag ( s )

The standard reduction potential can be determined by subtracting the standard reduction potential for the reaction occurring at the anode from the standard reduction potential for the reaction occurring at the cathode. The minus sign is needed because oxidation is the reverse of reduction.

E cell ° = E cathode ° E anode °
+0.80 V = E Ag + /Ag ° E H + /H 2 ° = E Ag + /Ag ° 0 = E Ag + /Ag °

It is important to note that the potential is not doubled for the cathode reaction.

The SHE is rather dangerous and rarely used in the laboratory. Its main significance is that it established the zero for standard reduction potentials. Once determined, standard reduction potentials can be used to determine the standard cell potential , E cell ° , for any cell. For example, for the cell shown in [link] ,

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
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bill
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bill
-24m+3+3mÁ^2
Susan
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Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
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Aphelele
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Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
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Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
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Grace
Send the example to me here and let me see
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A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
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Abubakar
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Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
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Method
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Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
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When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
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Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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