Standard Reduction (Electrode) Potentials at 25 °C Half-Cell Reaction E° (volts) F2(g) + 2 e→2 F(aq) 2.87 Ce4(aq) + e –→ Ce"(aq) 3+ 1.61 2+ MnO4 (aq) + 8 H"(aq) + 5 e¯ –→ Mn-T(aq) + 4 H2O(1) 1.51 Cl2(g) + 2 e' —2 СГ (aq) 1.36 Cr2072"(aq) + 14 H(aq) + 6 e¯ →2 Cr"(aq) + 7 H2O(1) 1.33 O2(g) + 4 H(aq) + 4 e→2 H2O(1) 1.229 Br2(1) + 2 e¯ –→2 Br¯(aq) 1.08 NO3 (aq) + 4 H"(aq) +3 e¯ –→ NO(g) + 2 H2O(1) 0.96 2 Hg-(aq) + 2 e¯ → Hg22 (aq) 0.920 Hg-"(aq) + 2 e →Hg(1) 0.855 Ag"(aq) + e¯ – Ag(s) 0.799 2+, Hg2"(aq) + 2 e¯ →2 Hg(1) 0.789 Fe(aq) + e → Fe2(aq) 0.771 I2(s) + 2 e → 2 1 (aq) 0.535 Fe(CN)6° (aq) + e→ Fe(CN)64 (aq) 0.48 Cu2(aq) + 2 e → Cu(s) 0.337 Cu2(aq) + e" – Cu"(aq) 0.153 S(s) + 2 H(aq) + 2 e → H2S(aq) 0.14 2 H(aq) + 2 e → H2(g) 0.0000 Pb2(aq) + 2 e → Pb(s) -0.126 Sn-"(aq) + 2 e¯ → Sn(s) „2+ -0.14 Ni2+(ag) + 2 e →Ni(s) -0.25 Co2+(ag) + 2 e →Co(s) -0.28 Cd2(aq) + 2 e →Cd(s) -0.403 Cr3+ C* (aq) (aq) + e -0.41 Fe2(aq) + 2 e →Fe(s) -0.44 Cr3+ (aq) + 3 e → Cr(s) -0.74 Zn2(aq) + 2 e →Zn(s) -0.763 2 H20(1) + 2 e → H2(g) + 2 OH (aq) -0.83 Mn-"(aq) + 2 e¯ → Mn(s) 2+ -1.18 Als(aq) + 3 e → Al(s) -1.66 2+ Mg"(aq) + 2 e →Mg(s) -2.37 Na"(aq) + e –→ Na(s) -2.714 K(aq) + e→ K(s) -2.925 Li(aq) + e Li(s) -3.045

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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The Nernst Equation

A non-standard cell or half-cell potential can be calculated using the Nernst Equation:

E = E^o- RT/nF (ln Q)

where

E	=	potential under non-standard conditions
E^o	=	standard potential
R	=	ideal gas constant
T	=	kelvin temperature
n	=	number of moles of electrons for the reaction as written
F	=	charge carried by 1 mol of electrons
Q	=	reaction quotient

It is customary to use the equation in a form where numerical values are substituted for R, T and F at a temperature of 25 °C.

For
	R	=	8.314 J mol^-1K^-1
	T	=	298.15 K
	F	=	96,485 J V^-1 mol^-1

RT/F = (8.314 J mol^-1 K^-1)(298.15 K)/((96,485 J V^-1 mol^-1)) = 0.0257 V

and the Nernst equation with the potentials in volts is:

E = E^o- (0.0257/n) ( ln Q)

nautral logarithm

Sometimes it is more convenient to use base-10 logarithms, and the substitution of 2.303 log for ln is made. (2.303 × 0.0257 = 0.0592) Then, the Nernst equation for base-10 logs at 25 °C is:

E = E^o- (0.0592/n)( log Q)

base-10 logarithm

A common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correct logarithm.

Standard Reduction (Electrode) Potentials at 25 °C
Half-Cell Reaction
E° (volts)
F2(g) + 2 e→2 F(aq)
2.87
Ce4*(aq) + e –→ Ce"(aq)
3+
1.61
2+
MnO4 (aq) + 8 H"(aq) + 5 e¯ –→ Mn-T(aq) + 4 H2O(1)
1.51
Cl2(g) + 2 e' —2 СГ (aq)
1.36
Cr2072"(aq) + 14 H*(aq) + 6 e¯
→2 Cr"(aq) + 7 H2O(1)
1.33
O2(g) + 4 H*(aq) + 4 e→2 H2O(1)
1.229
Br2(1) + 2 e¯ –→2 Br¯(aq)
1.08
NO3 (aq) + 4 H"(aq) +3 e¯ –→ NO(g) + 2 H2O(1)
0.96
2 Hg-*(aq) + 2 e¯ →
Hg22 (aq)
0.920
Hg-"(aq) + 2 e →Hg(1)
0.855
Ag"(aq) + e¯ – Ag(s)
0.799
2+,
Hg2"(aq) + 2 e¯ →2 Hg(1)
0.789
Fe*(aq) + e →
Fe2*(aq)
0.771
I2(s) + 2 e → 2 1 (aq)
0.535
Fe(CN)6° (aq) + e→ Fe(CN)64 (aq)
0.48
Cu2*(aq) + 2 e → Cu(s)
0.337
Cu2*(aq) + e" – Cu"(aq)
0.153
S(s) + 2 H*(aq) + 2 e → H2S(aq)
0.14
2 H*(aq) + 2 e → H2(g)
0.0000
Pb2*(aq) + 2 e → Pb(s)
-0.126
Sn-"(aq) + 2 e¯ → Sn(s)
„2+
-0.14
Ni2+(ag) + 2 e →Ni(s)
-0.25
Co2+(ag) + 2 e →Co(s)
-0.28
Cd2*(aq) + 2 e →Cd(s)
-0.403
Cr3+
C* (aq)
(aq) + e
-0.41
Fe2*(aq) + 2 e →Fe(s)
-0.44
Cr3+
(aq) + 3 e → Cr(s)
-0.74
Zn2*(aq) + 2 e →Zn(s)
-0.763
2 H20(1) + 2 e → H2(g) + 2 OH (aq)
-0.83
Mn-"(aq) + 2 e¯ → Mn(s)
2+
-1.18
Als*(aq) + 3 e → Al(s)
-1.66
2+
Mg"(aq) + 2 e →Mg(s)
-2.37
Na"(aq) + e –→ Na(s)
-2.714
K(aq) + e→ K(s)
-2.925
Li*(aq) + e Li(s)
-3.045

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