<< Chapter < Page
  Chemistry   Page 1 / 1
Chapter >> Page >
Standard Electrode (Half-Cell) Potentials
Half-Reaction E ° (V)
Ag + + e Ag +0.7996
AgCl + e Ag + Cl +0.22233
[ Ag ( CN ) 2 ] + e Ag + 2 CN −0.31
Ag 2 CrO 4 + 2 e 2 Ag + CrO 4 2− +0.45
[ Ag ( NH 3 ) 2 ] + + e Ag + 2 NH 3 +0.373
[ Ag ( S 2 O 3 ) 2 ] 3+ + e Ag + 2S 2 O 3 2− +0.017
[ AlF 6 ] 3− + 3 e Al + 6 F −2.07
Al 3+ + 3 e Al −1.662
Am 3+ + 3 e Am −2.048
Au 3+ + 3 e Au +1.498
Au + + e Au +1.692
Ba 2+ + 2 e Ba −2.912
Be 2+ + 2 e Be −1.847
Br 2 ( a q ) + 2 e 2 Br +1.0873
Ca 2+ + 2 e Ca −2.868
Ce 3 + 3 e Ce −2.483
Ce 4+ + e Ce 3+ +1.61
Cd 2+ + 2 e Cd −0.4030
[ Cd ( CN ) 4 ] 2− + 2 e Cd + 4 CN −1.09
[ Cd ( NH 3 ) 4 ] 2+ + 2 e Cd + 4 NH 3 −0.61
CdS + 2 e Cd + S 2− −1.17
Cl 2 + 2 e 2 Cl +1.35827
ClO 4 + H 2 O + 2 e ClO 3 + 2 OH +0.36
ClO 3 + H 2 O + 2 e ClO 2 + 2 OH +0.33
ClO 2 + H 2 O + 2 e ClO + 2 OH +0.66
ClO + H 2 O + 2 e Cl + 2 OH +0.89
ClO 4 + 2 H 3 O + + 2 e ClO 3 + 3 H 2 O +1.189
ClO 3 + 3 H 3 O + + 2 e HClO 2 + 4 H 2 O +1.21
HClO + H 3 O + + 2 e Cl + 2 H 2 O +1.482
HClO + H 3 O + + e 1 2 Cl 2 + 2 H 2 O +1.611
HClO 2 + 2 H 3 O + + 2 e HClO + 3 H 2 O +1.628
Co 3+ + e Co 2+ ( 2 mol // H 2 SO 4 ) +1.83
Co 2+ + 2 e Co −0.28
[ Co ( NH 3 ) 6 ] 3+ + e [ Co ( NH 3 ) 6 ] 2+ +0.1
Co ( OH ) 3 + e Co ( OH ) 2 + OH +0.17
Cr 3 + 3 e Cr −0.744
Cr 3+ + e Cr 2+ −0.407
Cr 2+ + 2 e Cr −0.913
[ Cu ( CN ) 2 ] + e Cu + 2 CN −0.43
CrO 4 2− + 4 H 2 O + 3 e Cr ( OH ) 3 + 5 OH −0.13
Cr 2 O 7 2− + 14 H 3 O + + 6 e 2 Cr 3+ + 21 H 2 O +1.232
[ Cr ( OH ) 4 ] + 3 e Cr + 4 OH −1.2
Cr ( OH ) 3 + 3 e Cr + 3 OH −1.48
Cu 2+ + e Cu + +0.153
Cu 2+ + 2 e Cu +0.34
Cu + + e Cu +0.521
F 2 + 2 e 2 F +2.866
Fe 2+ + 2 e Fe −0.447
Fe 3+ + e Fe 2+ +0.771
[ Fe ( CN ) 6 ] 3− + e [ Fe ( CN ) 6 ] 4− +0.36
Fe ( OH ) 2 + 2 e Fe + 2 OH −0.88
FeS + 2 e Fe + S 2− −1.01
Ga 3+ + 3 e Ga −0.549
Gd 3+ + 3 e Gd −2.279
1 2 H 2 + e H −2.23
2 H 2 O + 2 e H 2 + 2 OH −0.8277
H 2 O 2 + 2 H 3 O + + 2 e 4 H 2 O +1.776
2 H 3 O + + 2 e H 2 + 2 H 2 O 0.00
HO 2 + H 2 O + 2 e 3 OH +0.878
Hf 4+ + 4 e Hf −1.55
Hg 2+ + 2 e Hg +0.851
2 Hg 2+ + 2 e Hg 2 2+ +0.92
Hg 2 2+ + 2 e 2 Hg +0.7973
[ HgBr 4 ] 2− + 2 e Hg + 4 Br +0.21
Hg 2 Cl 2 + 2 e 2 Hg + 2 Cl +0.26808
[ Hg ( CN ) 4 ] 2− + 2e Hg + 4CN −0.37
[ HgI 4 ] 2− + 2e Hg + 4I −0.04
HgS + 2e Hg + S 2− −0.70
I 2 + 2e 2I +0.5355
In 3+ + 3e In −0.3382
K + + e K −2.931
La 3+ + 3e La −2.52
Li + + e Li −3.04
Lu 3+ + 3e Lu −2.28
Mg 2+ + 2e Mg −2.372
Mn 2+ + 2e Mn −1.185
MnO 2 + 2H 2 O + 2e Mn ( OH ) 2 + 2OH −0.05
MnO 4 + 2H 2 O + 3e MnO 2 + 4OH +0.558
MnO 2 + 4 H + + 2 e Mn 2+ + 2 H 2 O +1.23
MnO 4 + 8 H + + 5 e Mn 2+ + 4 H 2 O +1.507
Na + + e Na −2.71
Nd 3+ + 3e Nd −2.323
Ni 2+ + 2e Ni −0.257
[ Ni ( NH 3 ) 6 ] 2+ + 2e Ni + 6NH 3 −0.49
NiO 2 + 4 H + + 2 e Ni 2+ + 2 H 2 O +1.593
NiO 2 + 2H 2 O + 2e Ni ( OH ) 2 + 2OH +0.49
NiS + 2e Ni + S 2− +0.76
NO 3 + 4 H + + 3 e NO + 2 H 2 O +0.957
NO 3 + 3 H + + 2 e HNO 2 + H 2 O +0.92
NO 3 + H 2 O + 2e NO 2 + 2OH +0.10
Np 3+ + 3e Np −1.856
O 2 + 2H 2 O + 4e 4OH +0.401
O 2 + 2 H + + 2 e H 2 O 2 +0.695
O 2 + 4 H + + 4 e 2 H 2 O +1.229
Pb 2+ + 2e Pb −0.1262
PbO 2 + SO 4 2− + 4 H + + 2 e PbSO 4 + 2 H 2 O +1.69
PbS + 2e Pb + S 2− −0.95
PbSO 4 + 2e Pb + SO 4 2− −0.3505
Pd 2+ + 2e Pd +0.987
[ PdCl 4 ] 2− + 2e Pd + 4Cl +0.591
Pt 2+ + 2e Pt +1.20
[ PtBr 4 ] 2− + 2e Pt + 4Br +0.58
[ PtCl 4 ] 2− + 2e Pt + 4Cl +0.755
[ PtCl 6 ] 2− + 2e [ PtCl 4 ] 2− + 2Cl +0.68
Pu 3 + 3e Pu −2.03
Ra 2+ + 2e Ra −2.92
Rb + + e Rb −2.98
[ RhCl 6 ] 3− + 3e Rh + 6Cl +0.44
S + 2e S 2− −0.47627
S + 2 H + + 2 e H 2 S +0.142
Sc 3+ + 3e Sc −2.09
Se + 2 H + + 2 e H 2 Se −0.399
[ SiF 6 ] 2− + 4e Si + 6F −1.2
SiO 3 2− + 3H 2 O + 4e Si + 6OH −1.697
SiO 2 + 4 H + + 4 e Si + 2 H 2 O −0.86
Sm 3+ + 3e Sm −2.304
Sn 4+ + 2e Sn 2+ +0.151
Sn 2+ + 2e Sn −0.1375
[ SnF 6 ] 2− + 4e Sn + 6F −0.25
SnS + 2e Sn + S 2− −0.94
Sr 2+ + 2e Sr −2.89
TeO 2 + 4 H + + 4 e Te + 2 H 2 O +0.593
Th 4+ + 4e Th −1.90
Ti 2+ + 2e Ti −1.630
U 3+ + 3e U −1.79
V 2+ + 2e V −1.19
Y 3+ + 3e Y −2.37
Zn 2+ + 2e Zn −0.7618
[ Zn ( CN ) 4 ] 2− + 2e Zn + 4CN −1.26
[ Zn ( NH 3 ) 4 ] 2+ + 2e Zn + 4NH 3 −1.04
Zn ( OH ) 2 + 2e Zn + 2OH −1.245
[ Zn ( OH ) 4 ] 2 + 2e Zn + 4OH −1.199
ZnS + 2e Zn + S 2− −1.40
Zr 4 + 4e Zr −1.539

Questions & Answers

how do you get the 2/50
Abba Reply
number of sport play by 50 student construct discrete data
Aminu Reply
width of the frangebany leaves on how to write a introduction
Theresa Reply
Solve the mean of variance
Veronica Reply
Step 1: Find the mean. To find the mean, add up all the scores, then divide them by the number of scores. ... Step 2: Find each score's deviation from the mean. ... Step 3: Square each deviation from the mean. ... Step 4: Find the sum of squares. ... Step 5: Divide the sum of squares by n – 1 or N.
kenneth
what is error
Yakuba Reply
Is mistake done to something
Vutshila
Hy
anas
hy
What is the life teble
anas
hy
Jibrin
statistics is the analyzing of data
Tajudeen Reply
what is statics?
Zelalem Reply
how do you calculate mean
Gloria Reply
diveving the sum if all values
Shaynaynay
let A1,A2 and A3 events be independent,show that (A1)^c, (A2)^c and (A3)^c are independent?
Fisaye Reply
what is statistics
Akhisani Reply
data collected all over the world
Shaynaynay
construct a less than and more than table
Imad Reply
The sample of 16 students is taken. The average age in the sample was 22 years with astandard deviation of 6 years. Construct a 95% confidence interval for the age of the population.
Aschalew Reply
Bhartdarshan' is an internet-based travel agency wherein customer can see videos of the cities they plant to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400 a. what is the probability of getting more than 12,000 hits? b. what is the probability of getting fewer than 9,000 hits?
Akshay Reply
Bhartdarshan'is an internet-based travel agency wherein customer can see videos of the cities they plan to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400. a. What is the probability of getting more than 12,000 hits
Akshay
1
Bright
Sorry i want to learn more about this question
Bright
Someone help
Bright
a= 0.20233 b=0.3384
Sufiyan
a
Shaynaynay
How do I interpret level of significance?
Mohd Reply
It depends on your business problem or in Machine Learning you could use ROC- AUC cruve to decide the threshold value
Shivam
how skewness and kurtosis are used in statistics
Owen Reply
yes what is it
Taneeya
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

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