Wize University Chemistry Textbook > Electrochemistry

Important Equations + Nonstandard conditions (Nernst Equation)

Popular Courses

0:00 / 0:00

Cell Potentials and Gibbs Free Energy
 Free energy (ΔG°) and cell potential (E°cell) are related through the following equation:
∆G°=−nFE°cell∆G°=-nFE°_{cell}∆G°=−nFE°cell​
 ΔG° =  Free energy difference between products and reactants in their standard states 
 n      =  # moles of electrons
 F      =  Faraday constant  =96485Cmol e−=96485\frac{ C}{mol\ e^-}=96485mol e−C​
 E°cell =  Standard electrochemical reduction potential 
  ** If E°cell is positive, the process is spontaneous **
Remember that ΔG° and E° are for the cell at standard conditions (1 atm for gases and 1M for solutions)
∆G°=−nFE°cell=−RTln⁡⁡K∆G°=-nFE°_{cell}=-RT \ln⁡K∆G°=−nFE°cell​=−RTln⁡K

Ecello=RTnFln⁡KE_{cell}^o=\frac{RT}{nF}\ln KEcello​=nFRT​lnK

K=enFEoRTK=e^{\frac{nFE^o}{RT}}K=eRTnFEo​

Under non-standard conditions we have:
∆G=∆G°+RTln⁡⁡Q=−nFEcell∆G=∆G°+RT \ln⁡Q=-nFE_{cell}∆G=∆G°+RTln⁡Q=−nFEcell​
We can correct the E° to reflect the non-standard conditions by using the Nernst equation
Ecell=Ecell0−RTnFln⁡⁡Q or Ecell=Ecell0−0.05916Vnlog⁡⁡QE_{cell}=E_{cell}^0\cfrac{-RT}{nF}  \ln⁡Q   \    or \      E_{cell}=E_{cell}^0\cfrac{-0.05916V}{n}  \log⁡QEcell​=Ecell0​nF−RT​ln⁡Q or Ecell​=Ecell0​n−0.05916V​log⁡Q
R=gas constan⁡t (8.314JmolK)R=gas\ cons\tan t\ \left(\frac{8.314J}{molK}\right)R=gas constant (molK8.314J​) 
T=temperatureT =	temperatureT=temperature
    n=#of electronsn =	\# of\  electronsn=#of electrons
              F=Faradays ConstantF =	Faradays \ ConstantF=Faradays Constant
  

0:00 / 0:00

Calculate the deltaGo and Keq for following electrochemical cells: deltaGo= -nFEocell & Keq = enFEo/RT
Pb(s) + CuSO4(aq) --> Cu(s) + PbSO4(aq)  Temp 25 C

1) From a standard reduction potential table we can find the following: 
 Pb2+(aq) + 2e- -> Pb(s)     Eo=-0.13 V
 Cu2+(aq) + 2e- -> Cu(s)     Eo=0.34 V

2) Let's try to solve for deltaGo first:

deltaGo=-nFEocell 

We know n=moles of electrons involved
Here, n=2
 

F is just a constant (Faraday's constant): 96500C/mole- 

Eocell we can calculate by using the 2 half reactions, reversing a reaction based on the overall chemical reaction, and then adding up the Eo values for oxidation and reduction:
In our overall equation, when we look at the reactants, which is being oxidized and which is being reduced? 

Pb(s) is being oxidized, Cu2+ is being reduced.

Based on this info, which equation should we reverse from the table? First equation
What is the Eocell(redox)=? 

Eocell=0.13 + 0.34
Eocell=0.47 V 

Now we can plug these values in and solve for deltaGo: 

deltaGo=-nFEocell

             =(-2mole-)(96500C/mole-)(0.47V) 
             =-90710 J/mol 

If we wanted our deltaGo to be in kJ/mol what would our answer be? -90.71kJ/mol 

3) Now let's solve for Keq: 
Keq = enFEo/RT             

We already found n and Eo, we know F (96500C/mole-), we know R (8.314J/molK) and the temperature is given in the question (25C=298K!) Note: These equations are used for standard conditions, so the temperature would have to be 25C or 298K! 

Keq = enFEo/RT    

        =e(2)(96500)(0.47)/8.314x298
          =e36.61246
          =7.95x1015                         

Use the following reduction potentials to calculate the Ksp of La(OH)3(s)

La(OH)3(s)+3e−→La(s)+3 OH(aq)−Eo=−2.90VLa(aq)3++3e−→La(s)Eo=−2.379 V\begin{array}{llllll}&La(OH)_{3(s)}&+&3e-&\rightarrow& La_{(s)}&+&3\ OH^-_{(aq)}& &&E^o=-2.90 V \\ &La^{3+}_{(aq)}&+&3e^-&\rightarrow& La_{(s)}&&&&& E^o=-2.379\ V \end{array}​La(OH)3(s)​La(aq)3+​​++​3e−3e−​→→​La(s)​La(s)​​+​3 OH(aq)−​​​​Eo=−2.90VEo=−2.379 V​

3.64×10−273.64\times 10^{-27}3.64×10−27

3.56×10−273.56\times 10^{-27}3.56×10−27

3.67×10−273.67\times 10^{-27}3.67×10−27

3.54×10−273.54\times 10^{-27}3.54×10−27

I don't know

0:00 / 0:00

Cells Operating at Non-Standard Conditions: The Nernst Equation 
 If we want to operate a cell at non-standard conditions (not 1 M solutions) we need to use the Nernst equation to find new electromotive force 
Ecell=Ecello−RTnFln⁡QE_{cell}=E^o_{cell}-\cfrac{RT}{nF}\ln QEcell​=Ecello​−nFRT​lnQ
 R: The ideal gas constant (8.314J/molK)
 T: Temperature (K)
 n: # of electrons involved 
 F: The Faraday Constant (96500 C/mol e-)
Eocell: standard electromotive force 

0:00 / 0:00

Determine the voltage produced in a galvanic cell at room temperature made up of the following half reactions at a pH of 3.5 and the following concentrations. [MnO4 - ] = 0.5 M, [Mn2+] = 0.4 M, [Sn4+] = 2.0 M and [Sn2+] = 0.8 M R=8.314J/molK 
                                    Sn(aq)4++2e−→Sn(aq)2+Eo=0.15VSn^{4+}_{(aq)} +2e^- \to Sn^{2+}_{(aq)} \hspace{110pt} E^o=0.15VSn(aq)4+​+2e−→Sn(aq)2+​Eo=0.15V
                                  MnO4(aq)−+5e−+8H(aq)+→Mn(aq)2++4H2O(l)Eo=1.49VMnO_{4(aq)}^- +5e^-+8H^+_{(aq)} \to Mn^{2+}_{(aq)} +4H_2O_{(l)} \hspace{10pt} E^o=1.49VMnO4(aq)−​+5e−+8H(aq)+​→Mn(aq)2+​+4H2​O(l)​Eo=1.49V

 A Galvanic cell must have a positive voltage so the Sn half-cell must be the oxidation half cell. 
E=1.49V–0.15V=1.34VE = 1.49 V – 0.15 V = 1.34 VE=1.49V–0.15V=1.34V
 The chemical equation id shown below: 
5Sn(aq)2++2MnO(aq)4−+16H(aq)+→5Sn(aq)4++2Mn(aq)2++8H2O(l)5Sn_{(aq)}^{2+}+2MnO_{(aq)}^{4-}+16H_{(aq)}^+→5Sn_{(aq)}^{4+}+2Mn_{(aq)}^{2+}+8H_2O_{(l)}5Sn(aq)2+​+2MnO(aq)4−​+16H(aq)+​→5Sn(aq)4+​+2Mn(aq)2+​+8H2​O(l)​

 
Using the Nernst Equation: 
Ecell=Ecello−RTnFln⁡QE_{cell}=E^o_{cell}-\cfrac{RT}{nF}\ln QEcell​=Ecello​−nFRT​lnQ

      
=1.34V−(8.314JK−1mol−1)(298K)(10)(96485Cmol−1)ln⁡([2]5[0.4]2[0.8]5[0.5]2[10−3.5]16)=1.34V-\cfrac{(8.314JK^{-1}mol^{-1})(298K)}{(10)(96485Cmol^{-1})}\ln\bigg(\cfrac{[2]^{5}[0.4]^2}{[0.8]^{5}[0.5]^2[10^{-3.5}]^{16}}\bigg)=1.34V−(10)(96485Cmol−1)(8.314JK−1mol−1)(298K)​ln([0.8]5[0.5]2[10−3.5]16[2]5[0.4]2​)

=-340V

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

At 25C, what is the potential of the following cell and what is the Keq?
Ni|Ni2+(aq)(0.010 M)||Cr-(0.20 M)|Cr2(g) | Pt

1) Is the reaction occuring in standard or non-standard conditions?
If it is occuring in standard conditions --> it is asking us to solve for Eocell 
If it is occuring in nonstandard conditions --> it is asking us to solve for Ecell 


PAGE BREAK

2) Which equation will we use to solve this problem? 
Nernst equation for non-standard conditions!
E = Eo - (RT/nF)ln(Q)

E is what the question is asking us to solve 
Eo we can calculate by using the SRP from a table 
R=8.314J/molK
n=# of electrons involved in the overall reaction
F=faraday's constant (96500C/mole-)
Q=[products]/[reactants]

3) Solve for Eocell using SRP tables: 
Ni2+(aq) + 2e- --> Ni(s)   Eo=-0.25 V
Cr2(g) + 2e- --> 2Cr-(aq)   Eo=+1.36 V

From this the standard cell reaction would be:
Cr2(g) + 2e- --> 2Cr-(aq)
Eo=
Ni(s) --> Ni2+(aq) + 2e-
Eo=
to give Cr2(g) + Ni(s) --> 2Cr-(aq) + Ni2+(aq)
Eo=
1.61V

4) Now we can determine E at non-standard conditions via the NERNST Equation:
E = Eo - (RT/nF)ln(Q) 

Bonus: Is this reaction spontaneous or not? 

5) What is Keq at 298 K?
Keq = enFEo/RT

I have answered this question

Extra Practice

A comproportionation reaction is a reaction where one element in two different oxidation states reacts to form a product where both atoms now have the same oxidation state. An electrochemical cell can be constructed using the comproportionation of cobalt as shown below. 
CO3++CO+→2CO2+CO^{3+}+CO^+ \to 2CO^{2+}CO3++CO+→2CO2+
If the equilibrium constant for the above reaction is found to be 6.3 x 104 determine the standard cell potential ε0 and draw the electrochemical cell in standard notation 

The E° for the reaction of Cu/Cu2+ and Sn/Sn2+ is + 0.19 V. What is the value of the equilibrium constant K for this reaction at 298 K?

Calculate the standard voltage and write the standard cell notation for a galvanic cell composed of the following two half reactions.
Fe2+ + 2e-  --> Fe(s)  Eo=-0.41V 
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O  Eo=1.33V 

Electrochemistry: Galvanic cells

For each galvanic cell, what is the value of the equilibrium constant for the reaction between each pair at 25 °C?  Please ensure that your answer is specified to two decimal places and that your answer is formatted as follows: 9.15x1025 for both answers.
 Ni(s) and Ag+(aq) 
 Fe(s) and Cr3+(aq) 

If you could increase the concentration of Cu2+, which of the following would be true about the cell potential? 

You make a cell with a copper electrode in a solution of copper nitrate and a silver electrode in a solution of silver nitrate.  
Cu(s)+2Ag+(aq)⇌Cu2+(aq)+2Ag(s)      E=0.46VCu(s)+2Ag^+(aq)⇌Cu^{2+}(aq)+2Ag(s)\ \ \ \ \ \ E = 0.46 VCu(s)+2Ag+(aq)⇌Cu2+(aq)+2Ag(s)      E=0.46V
If you could increase the concentration of Ag+ , which of the following would be true about the cell potential? 

Electrochemistry Additional Practice

At 25C, what is the potential of the following cell and what is the Keq?
Ni|Ni2+(aq)(0.010 M)||Cr-(0.20 M)|Cr2(g) | Pt
1) Is the reaction occuring in standard or non-standard conditions?

Gibbs Free Energy and Voltage Practice Problem #1

What is the equilibrium constant for the reaction of copper metal with bromine to form copper(II) ions and bromide ions in aqueous solution at 25oC?
Cu(aq)2++2e−→Cu(s)Eo=+0.34 VBr2(aq)+2e−→2 Br(aq)−Eo=+1.07 V\begin{array}{cc}
Cu^{2+}_{(aq)}+2e^- \rightarrow Cu_{(s)} & E^o=+0.34\ V\\
Br_{2(aq)}+2e^- \rightarrow 2\ Br^-_{(aq)} & E^o=+1.07\ V
\end{array}Cu(aq)2+​+2e−→Cu(s)​Br2(aq)​+2e−→2 Br(aq)−​​Eo=+0.34 VEo=+1.07 V​