Calculate the cell potential for the reaction as written at 25.00 °C , given that [Mg2+]=0.891 M and [Fe2+]=0.0100 M . Use the standard reduction potentials in this table.
Mg(s)+Fe2+(aq)↽−−⇀ Mg2+(aq)+Fe(s)
Can someone help me figure out step by step what I should do? I know that I need to find the standard E value, but I'm not quite sure which number I should reverse the sign for. I am also having trouble with identifying what the n value is.
Sapling #15 (Week 7/8)
Moderators: Chem_Mod, Chem_Admin
Re: Sapling #15 (Week 7/8)
This is the solution from sapling, hope it helps!
Use the Nernst equation to find the cell potential under these conditions
E=E0−RTnFlnQ
where E is the cell potential, E0 is the standard cell potential, R is the gas contant (8.3145 J/(mol⋅K)) , T is the Kelvin temperature, n is the number of electrons transferred, F is the Faraday constant (96485 J/(V⋅mol)) , and Q is the reaction quotient.
First, write the half-reactions. The reduction half-reaction occurs at the cathode and the oxidation half-reaction occurs at the anode.
Anode:Cr (s)⟶ Cr2+(aq)+2e−
Cathode:Fe2+(aq)+2e−⟶ Fe (s)
Use a table of standard reduction potentials to determine the reduction potential of each half-reaction and the overall standard cell potential.
E0=E0cathode−E0anode=(−0.44 V)−(−0.91 V)=0.47 V
Two moles of electrons are transferred per mole of reaction, so n=2 .
Convert the temperature to kelvins.
T=25.00 °C+273.15=298.15 K
The reaction quotient, Q , is the concentrations of products raised to their coefficients, divided by the concentrations of reactants raised to their coefficients.
Q=[Cr2+][Fe2+]=(0.823 M)(0.0100 M)=82.3
Insert these numbers into the Nernst equation to find the cell potential.
E=E0-RTnFlnQ
E=(0.47 V)−(8.3145Jmol⋅K)(298.15 K)(2)(96485JV⋅mol)ln(82.3)=0.41 V
Use the Nernst equation to find the cell potential under these conditions
E=E0−RTnFlnQ
where E is the cell potential, E0 is the standard cell potential, R is the gas contant (8.3145 J/(mol⋅K)) , T is the Kelvin temperature, n is the number of electrons transferred, F is the Faraday constant (96485 J/(V⋅mol)) , and Q is the reaction quotient.
First, write the half-reactions. The reduction half-reaction occurs at the cathode and the oxidation half-reaction occurs at the anode.
Anode:Cr (s)⟶ Cr2+(aq)+2e−
Cathode:Fe2+(aq)+2e−⟶ Fe (s)
Use a table of standard reduction potentials to determine the reduction potential of each half-reaction and the overall standard cell potential.
E0=E0cathode−E0anode=(−0.44 V)−(−0.91 V)=0.47 V
Two moles of electrons are transferred per mole of reaction, so n=2 .
Convert the temperature to kelvins.
T=25.00 °C+273.15=298.15 K
The reaction quotient, Q , is the concentrations of products raised to their coefficients, divided by the concentrations of reactants raised to their coefficients.
Q=[Cr2+][Fe2+]=(0.823 M)(0.0100 M)=82.3
Insert these numbers into the Nernst equation to find the cell potential.
E=E0-RTnFlnQ
E=(0.47 V)−(8.3145Jmol⋅K)(298.15 K)(2)(96485JV⋅mol)ln(82.3)=0.41 V
Re: Sapling #15 (Week 7/8)
If one of the 1/2 RXNs transferred 2 mol of reaction for one product and the other transferred 3, would we say that the overall moles of electrons transferred in the reaction was 6?
Re: Sapling #15 (Week 7/8)
Or, would you add them up and say that 5 moles of electrons were transferred?
-
Ally Mosher
- Posts: 113
- Joined: Fri Sep 24, 2021 5:26 am
-
Kathryn Heinemeier 3H
- Posts: 104
- Joined: Fri Sep 24, 2021 5:09 am
Re: Sapling #15 (Week 7/8)
it would be 6 electrons because you find the common denominator between the two essentially
Who is online
Users browsing this forum: No registered users and 5 guests