<< Chapter < Page Chapter >> Page >
Cu ( s ) Cu 2+ ( a q , 1 M ) Ag + ( a q , 1 M ) Ag ( s )
anode (oxidation): Cu ( s ) Cu 2+ ( a q ) + 2e cathode (reduction): 2 Ag + ( a q ) + 2e 2Ag ( s ) ¯ overall: Cu ( s ) + 2Ag + ( a q ) Cu 2+ ( a q ) + 2Ag ( s )
E cell ° = E cathode ° E anode ° = E Ag + /Ag ° E Cu 2+ /Cu ° = 0.80 V 0.34 V = 0.4 6 V

Again, note that when calculating E cell ° , standard reduction potentials always remain the same even when a half-reaction is multiplied by a factor. Standard reduction potentials for selected reduction reactions are shown in [link] . A more complete list is provided in Appendix L .

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 which is labeled “H N O subscript 3 ( a q ).” The beaker on the right contains a clear, colorless solution which is labeled “A g N O subscript 3 ( a q ).” A glass tube in the shape of an inverted U connects the two beakers at the center of the diagram and is labeled “Salt bridge.” 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 label “2 N a superscript plus” appears on the upper right portion of the tube. A curved arrow extends from this label down and to the right. The label “2 N O subscript 3 superscript negative” appears on the upper left portion of the tube. A curved arrow extends from this label down and to the left. The beaker on the left has a glass tube partially submerged in the liquid. Bubbles are rising from the grey square, labeled “SHE anode” at the bottom of the tube. A curved arrow points up to the right. The labels “2 H superscript plus” and “2 N O subscript 3 superscript negative” appear on the liquid in the beaker. 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 0.80 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 a silver strip that is labeled “A g cathode.” 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. The solution in the beaker on the right has the labels “N O subscript 3 superscript negative” and “A g superscript plus” on the solution. A curved arrow extends from the A g superscript plus label to the A g cathode. Below the left beaker at the bottom of the diagram is the label “Oxidation half-reaction: H subscript 2 ( g ) right pointing arrow 2 H superscript plus ( a q ) plus 2 e superscript negative.” Below the right beaker at the bottom of the diagram is the label “Reduction half-reaction: 2 A g superscript plus ( a q ) right pointing arrow 2 A g ( s ).”
A galvanic cell can be used to determine the standard reduction potential of Ag + . The SHE on the left is the anode and assigned a standard reduction potential of zero.
Selected Standard Reduction Potentials at 25 °C
Half-Reaction E ° (V)
F 2 ( g ) + 2e 2F ( a q ) +2.866
PbO 2 ( s ) + SO 4 2− ( a q ) + 4H + ( a q ) + 2e PbSO 4 ( s ) + 2H 2 O ( l ) +1.69
MnO 4 ( a q ) + 8H + ( a q ) + 5e Mn 2+ ( a q ) + 4H 2 O ( l ) +1.507
Au 3+ ( a q ) + 3e Au ( s ) +1.498
Cl 2 ( g ) + 2e 2Cl ( a q ) +1.35827
O 2 ( g ) + 4H + ( a q ) + 4e 2H 2 O ( l ) +1.229
Pt 2+ ( a q ) + 2e Pt ( s ) +1.20
Br 2 ( a q ) + 2e 2Br ( a q ) +1.0873
Ag + ( a q ) + e Ag ( s ) +0.7996
Hg 2 2+ ( a q ) + 2e 2Hg ( l ) +0.7973
Fe 3+ ( a q ) + e Fe 2+ ( a q ) +0.771
MnO 4 ( a q ) + 2H 2 O ( l ) + 3e MnO 2 ( s ) + 4OH ( a q ) +0.558
I 2 ( s ) + 2e 2I ( a q ) +0.5355
NiO 2 ( s ) + 2H 2 O ( l ) + 2e Ni(OH) 2 ( s ) + 2OH ( a q ) +0.49
Cu 2+ ( a q ) + 2e Cu ( s ) +0.337
Hg 2 Cl 2 ( s ) + 2e 2Hg ( l ) + 2Cl ( a q ) +0.26808
AgCl ( s ) + 2e Ag ( s ) + Cl ( a q ) +0.22233
Sn 4+ ( a q ) + 2e Sn 2+ ( a q ) +0.151
2H + ( a q ) + 2e H 2 ( g ) 0.00
Pb 2+ ( a q ) + 2e Pb ( s ) −0.126
Sn 2+ ( a q ) + 2e Sn ( s ) −0.1262
Ni 2+ ( a q ) + 2e Ni ( s ) −0.257
Co 2+ ( a q ) + 2e Co ( s ) −0.28
PbSO 4 ( s ) + 2e Pb ( s ) + SO 4 2− ( a q ) −0.3505
Cd 2+ ( a q ) + 2e Cd ( s ) −0.4030
Fe 2+ ( a q ) + 2e Fe ( s ) −0.447
Cr 3+ ( a q ) + 3e Cr ( s ) −0.744
Mn 2+ ( a q ) + 2e Mn ( s ) −1.185
Zn(OH) 2 ( s ) + 2e Zn ( s ) + 2OH ( a q ) −1.245
Zn 2+ ( a q ) + 2e Zn ( s ) −0.7618
Al 3+ ( a q ) + 3e Al ( s ) −1.662
Mg 2 ( a q ) + 2e Mg ( s ) −2.372
Na + ( a q ) + e Na ( s ) −2.71
Ca 2+ ( a q ) + 2e Ca ( s ) −2.868
Ba 2+ ( a q ) + 2e Ba ( s ) −2.912
K + ( a q ) + e K ( s ) −2.931
Li + ( a q ) + e Li ( s ) −3.04

Tables like this make it possible to determine the standard cell potential for many oxidation-reduction reactions.

Cell potentials from standard reduction potentials

What is the standard cell potential for a galvanic cell that consists of Au 3+ /Au and Ni 2+ /Ni half-cells? Identify the oxidizing and reducing agents.

Solution

Using [link] , the reactions involved in the galvanic cell, both written as reductions, are

Au 3+ ( a q ) + 3 e Au ( s ) E Au 3+ /Au ° = +1.498 V
Ni 2+ ( a q ) + 2 e Ni ( s ) E Ni 2+ /Ni ° = −0.257 V

Galvanic cells have positive cell potentials, and all the reduction reactions are reversible. The reaction at the anode will be the half-reaction with the smaller or more negative standard reduction potential. Reversing the reaction at the anode (to show the oxidation) but not its standard reduction potential gives:

Anode (oxidation): Ni ( s ) Ni 2+ ( a q ) + 2e E anode ° = E Ni 2+ /Ni ° = −0.257 V Cathode (reduction): Au 3+ ( a q ) + 3e Au ( s ) E cathode ° = E Au 3+ /Au ° = +1.498 V

The least common factor is six, so the overall reaction is

3Ni ( s ) + 2Au 3+ ( a q ) 3Ni 2+ ( a q ) + 2Au ( s )

The reduction potentials are not scaled by the stoichiometric coefficients when calculating the cell potential, and the unmodified standard reduction potentials must be used.

E cell ° = E cathode ° E anode ° = 1.498 V ( −0.2 57 V ) = 1.7 55 V

From the half-reactions, Ni is oxidized, so it is the reducing agent, and Au 3+ is reduced, so it is the oxidizing agent.

Check your learning

A galvanic cell consists of a Mg electrode in 1 M Mg(NO 3 ) 2 solution and a Ag electrode in 1 M AgNO 3 solution. Calculate the standard cell potential at 25 °C.

Answer:

Mg ( s ) + 2 Ag + ( a q ) Mg 2+ ( a q ) + 2 Ag ( s ) E cell ° = 0.7 996 V ( −2.3 72 V ) = 3.17 2 V

Got questions? Get instant answers now!

Key concepts and summary

Assigning the potential of the standard hydrogen electrode (SHE) as zero volts allows the determination of standard reduction potentials, , for half-reactions in electrochemical cells. As the name implies, standard reduction potentials use standard states (1 bar or 1 atm for gases; 1 M for solutes, often at 298.15 K) and are written as reductions (where electrons appear on the left side of the equation). The reduction reactions are reversible, so standard cell potentials can be calculated by subtracting the standard reduction potential for the reaction at the anode from the standard reduction for the reaction at the cathode. When calculating the standard cell potential, the standard reduction potentials are not scaled by the stoichiometric coefficients in the balanced overall equation.

Key equations

  • E cell ° = E cathode ° E anode °

Chemistry end of chapter exercises

For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

(a) Mg ( s ) + Ni 2+ ( a q ) Mg 2+ ( a q ) + Ni ( s )

(b) 2 Ag + ( a q ) + Cu ( s ) Cu 2+ ( a q ) + 2Ag ( s )

(c) Mn ( s ) + Sn(NO 3 ) 2 ( a q ) Mn(NO 3 ) 2 ( a q ) + Sn ( s )

(d) 3 Fe(NO 3 ) 2 ( a q ) + Au(NO 3 ) 3 ( a q ) 3Fe(NO 3 ) 3 ( a q ) + Au ( s )

(a) +2.115 V (spontaneous); (b) +0.4626 V (spontaneous); (c) +1.0589 V (spontaneous); (d) +0.727 V (spontaneous)

Got questions? Get instant answers now!

For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

(a) Mn ( s ) + Ni 2+ ( a q ) Mn 2+ ( a q ) + Ni ( s )

(b) 3 Cu 2+ ( a q ) + 2Al ( s ) 2Al 3+ ( a q ) + 2Cu ( s )

(c) Na ( s ) + LiNO 3 ( a q ) NaNO 3 ( a q ) + Li ( s )

(d) Ca(NO 3 ) 2 ( a q ) + Ba ( s ) Ba(NO 3 ) 2 ( a q ) + Ca ( s )

Got questions? Get instant answers now!

Determine the overall reaction and its standard cell potential at 25 °C for this reaction. Is the reaction spontaneous at standard conditions?

Cu ( s ) Cu 2+ ( a q ) Au 3+ ( a q ) Au ( s )

3 Cu ( s ) + 2Au 3+ ( a q ) 3Cu 2+ ( a q ) + 2Au ( s ) ; +1.16 V; spontaneous

Got questions? Get instant answers now!

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell made from a half-cell consisting of a silver electrode in 1 M silver nitrate solution and a half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?

Got questions? Get instant answers now!

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell in which cadmium metal is oxidized to 1 M cadmium(II) ion and a half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?

3 Cd ( s ) + 2Al 3+ ( a q ) 3Cd 2+ ( a q ) + 2Al ( s ) ; −1.259 V; nonspontaneous

Got questions? Get instant answers now!

Determine the overall reaction and its standard cell potential at 25 °C for these reactions. Is the reaction spontaneous at standard conditions? Assume the standard reduction for Br 2 ( l ) is the same as for Br 2 ( aq ).
Pt ( s ) H 2 ( g ) H + ( a q ) Br 2 ( a q ) Br ( a q ) Pt ( s )

Got questions? Get instant answers now!

Questions & Answers

it is the relatively stable flow of income
Chidubem Reply
what is circular flow of income
Divine Reply
branches of macroeconomics
SHEDRACK Reply
what is Flexible exchang rate?
poudel Reply
is gdp a reliable measurement of wealth
Atega Reply
introduction to econometrics
Husseini Reply
Hi
mostafa
hi
LEMLEM
hello
Sammol
hi
Mahesh
bi
Ruqayat
hi
Ruqayat
Hi fellas
Nyawa
hey
Sammol
hi
God
hello
Jahara
Good morning
Jorge
hi
abubakar
hi
Nmesoma
hi
Mahesh
Hi
Tom
Why is unemployment rate never zero at full employment?
Priyanka Reply
bcoz of existence of frictional unemployment in our economy.
Umashankar
what is flexible exchang rate?
poudel
due to existence of the pple with disabilities
Abdulraufu
the demand of a good rises, causing the demand for another good to fall
Rushawn Reply
is it possible to leave every good at the same level
Joseph
I don't think so. because check it, if the demand for chicken increases, people will no longer consume fish like they used to causing a fall in the demand for fish
Anuolu
is not really possible to let the value of a goods to be same at the same time.....
Salome
Suppose the inflation rate is 6%, does it mean that all the goods you purchase will cost 6% more than previous year? Provide with reasoning.
Geetha Reply
Not necessarily. To measure the inflation rate economists normally use an averaged price index of a basket of certain goods. So if you purchase goods included in the basket, you will notice that you pay 6% more, otherwise not necessarily.
Waeth
discus major problems of macroeconomics
Alii Reply
what is the problem of macroeconomics
Yoal
Economic growth Stable prices and low unemployment
Ephraim
explain inflationcause and itis degre
Miresa Reply
what is inflation
Getu
increase in general price levels
WEETO
Good day How do I calculate this question: C= 100+5yd G= 2000 T= 2000 I(planned)=200. Suppose the actual output is 3000. What is the level of planned expenditures at this level of output?
Chisomo Reply
how to calculate actual output?
Chisomo
how to calculate the equilibrium income
Beshir
Criteria for determining money supply
Thapase Reply
who we can define macroeconomics in one line
Muhammad
Aggregate demand
Mohammed
C=k100 +9y and i=k50.calculate the equilibrium level of output
Mercy Reply
Hi
Isiaka
Hi
Geli
hy
Man
👋
Bahunda
hy how are you?
Man
ys
Amisha
how are you guys
Sekou
f9 guys
Amisha
how are you guys
Sekou
ys am also fine
Amisha
fine and you guys
Geli
from Nepal
Amisha
nawalparasi district from belatari
Amisha
nd u
Amisha
I am Camara from Guinea west Africa... happy to meet you guys here
Sekou
ma management ho
Amisha
ahile becheclor ho
Amisha
hjr ktm bta ho ani k kaam grnu hunxa tw
Amisha
belatari
Amisha
1st year ho
Amisha
nd u
Amisha
ahh
Amisha
kaha biratnagar
Amisha
ys
Amisha
kina k vo
Amisha
money as unit of account means what?
Kalombe
A unit of account is something that can be used to value goods and services and make calculations
Jim
all of you please speak in English I can't understand you're language
Muhammad
I want to know how can we define macroeconomics in one line
Muhammad
it must be .9 or 0.9 no Mpc is greater than 1 Y=100+.9Y+50 Y-.9Y=150 0.1Y/0.1=150/0.1 Y=1500
Kalombe
Mercy is it clear?😋
Kalombe
hi can someone help me on this question If a negative shocks shifts the IS curve to the left, what type of policy do you suggest so as to stabilize the level of output? discuss your answer using appropriate graph.
Galge Reply
if interest rate is increased this will will reduce the level of income shifting the curve to the left ◀️
Kalombe
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 3

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Chemistry' conversation and receive update notifications?

Ask