For the following electrochemical cell, Co(s)/Co2(aq,.0155M)// Ag+(aq,2.50M)/Ag(s)
1.Write the net cell equation.
2.Calculate the following values at 25 degrees celcius. Eocell, Ecell, deltaGo rxn, and delta Grxn
1. Write the net cell equation:
Half-reactions:
Co²⁺(aq) + 2e⁻ → Co(s) (Cobalt reduction)
Ag⁺(aq) + e⁻ → Ag(s) (Siver reduction)
Multiplying the second reaction by 2 to make the number of electrons equal:
Co²⁺(aq) + 2e⁻ → Co(s)
2(Ag⁺(aq) + e⁻ → Ag(s))
Adding the two half-reactions:
Co²⁺(aq) + 2Ag⁺(aq) → Co(s) + 2Ag(s)
2. Calculate the following values at 25 degrees Celsius. Eocell, Ecell, deltaGo rxn, and delta Grxn:
E0cell is the difference in standard reduction potential of the two half-reactions. Using the standard reduction potentials (look up from a table):
E0(Co²⁺/Co) = -0.28 V
E0(Ag⁺/Ag) = +0.80 V
E0cell = E0(Ag⁺/Ag) - E0(Co²⁺/Co) = 0.80 V - (-0.28 V) = 1.08 V
To calculate Ecell, use the Nernst equation:
Ecell = E0cell - (RT/nF) * ln(Q)
Where R is the gas constant = 8.314 J/(mol K), T is the temperature in Kelvin (25 degrees Celsius = 298 K), n is the number of moles of electrons transferred (2 in this case), F is Faraday's constant = 96,485 C/mol and Q is the reaction quotient, given by:
Q = [Co²⁺]/[Ag⁺]^2 = 0.0155 M / (2.50 M)^2
Ecell = 1.08 V - (8.314 J/(mol K) * 298 K / (2 * 96485 C/mol)) * ln(0.0155 / (2.5^2))
Ecell ≈ 1.08V - 0.0120 V = 1.068 V
To calculate deltaGo rxn (the standard Gibbs free energy change), use the equation:
deltaGo rxn = - nFE0cell
deltaGo rxn = -(2 mol e⁻)(96485 C/mol)(1.08 V)
deltaGo rxn ≈ -208 kJ/mol
To calculate delta Grxn (the Gibbs free energy change under nonstandard conditions), use the equation:
delta Grxn = - nFEcell
delta Grxn = -(2 mol e⁻)(96485 C/mol)(1.068 V)
delta Grxn ≈ -206.8 kJ/mol
So, the values for the given electrochemical cell at 25 degrees Celsius are:
E0cell = 1.08 V
Ecell = 1.068 V
deltaGo rxn = -208 kJ/mol
delta Grxn = -206.8 kJ/mol
To write the net cell equation, you need to identify the oxidation and reduction half-reactions involved in the given electrochemical cell.
1. Oxidation Half-Reaction:
Co(s) → Co2+(aq) + 2e-
2. Reduction Half-Reaction:
Ag+(aq) + e- → Ag(s)
To balance the number of electrons transferred in both half-reactions, the oxidation half-reaction needs to be multiplied by 2:
2Co(s) → 2Co2+(aq) + 4e-
Now, the net cell equation can be written by adding the oxidation and reduction half-reactions:
2Co(s) + 2Ag+(aq) → 2Co2+(aq) + 2Ag(s)
To calculate the following values at 25 degrees Celsius, you need to use the Nernst equation and standard electrode potentials (Eo) for the given half-reactions:
1. Eocell:
Eocell = Eo(reduction) - Eo(oxidation)
The standard electrode potentials for the half-reactions can be found in reference tables (e.g., a standard reduction potential table). Look up the values for Ag+(aq)/Ag(s) and Co2+(aq)/Co(s) half-reactions, subtract the Eo(oxidation) from the Eo(reduction), and calculate Eocell.
2. Ecell:
Ecell = Eocell - (0.05916/n) * log(Q)
Here, n is the number of electrons transferred in the balanced net cell equation, and Q is the reaction quotient. The reaction quotient can be calculated using concentrations of reactants and products involved in the cell equation.
3. deltaGo rxn:
deltaGo rxn = -nF * Eocell
Here, F is the Faraday constant (96,485 C/mol) and n is the number of moles of electrons transferred in the balanced net cell equation.
4. deltaGrxn:
deltaGrxn = deltaGo rxn + RT * ln(Q)
Here, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and Q is the reaction quotient.
By plugging in the appropriate values into the equations, you can calculate Eocell, Ecell, deltaGo rxn, and deltaGrxn at 25 degrees Celsius.
To write the net cell equation, we need to identify the half-reactions occurring at the anode and the cathode.
Anode (Oxidation): Co(s) -> Co2+(aq) + 2e^-
Cathode (Reduction): Ag+(aq) + e^- -> Ag(s)
To balance the half-reactions, we need to multiply the coefficients as necessary. In this case, we don't need to balance anything since the number of electrons involved in each half-reaction is already balanced.
Net Cell Equation:
Co(s) + Ag+(aq) -> Co2+(aq) + Ag(s)
Now, let's move on to calculating the values at 25 degrees Celsius.
1. Eocell (Standard Cell Potential):
We can calculate the standard cell potential using the standard reduction potentials (Eo) for each half-reaction. The standard reduction potential for the Co2+/Co couple is +0.28 V, while the standard reduction potential for the Ag+/Ag couple is +0.80 V.
Eocell = Eored(cathode) - Eored(anode)
= (+0.80 V) - (+0.28 V)
= +0.52 V
2. Ecell (Cell Potential):
The cell potential (Ecell) depends on the actual concentration of the species involved in the cell, using the Nernst equation:
Ecell = Eocell - (0.0592/n) * log(Q)
Where:
- Ecell is the cell potential
- Eocell is the standard cell potential
- n is the number of moles of electrons transferred
- Q is the reaction quotient
Since the stoichiometric coefficient for electrons in these half-reactions is 1, n = 1.
At 25 degrees Celsius, the reaction quotient is equal to the concentration of products divided by the concentration of reactants.
Q = [Co2+]/[Ag+]
Given concentrations:
[Co2+] = 0.0155 M
[Ag+] = 2.50 M
Q = (0.0155 M) / (2.50 M)
Q = 0.0062
Now we can substitute the values into the Nernst equation to calculate Ecell:
Ecell = Eocell - (0.0592/n) * log(Q)
= 0.52 V - (0.0592/1) * log(0.0062)
= 0.52 V - (0.0592) * (-2.21)
= 0.52 V + 0.1305
= 0.6505 V
3. ΔGorxn (Standard Gibbs Free Energy Change):
ΔGorxn = -n * F * Eocell
Where:
- ΔGorxn is the standard Gibbs free energy change
- n is the number of moles of electrons transferred
- F is Faraday's constant (96485 C/mol)
With n = 1 and F = 96485 C/mol, we can compute:
ΔGorxn = -(1 mol) * (96485 C/mol) * (0.52 V)
= -50199 C
Now, we can convert C (coulombs) to Joules (J):
1 C = 1 J
ΔGorxn = -50199 J
4. ΔGrxn (Gibbs Free Energy Change):
ΔGrxn = ΔGorxn + (RT * ln(Q))
Where:
- ΔGrxn is the actual Gibbs free energy change
- ΔGorxn is the standard Gibbs free energy change
- R is the ideal gas constant (8.314 J/(mol K))
- T is the temperature in Kelvin
Given that the temperature is 25°C = 298.15 K, we can calculate:
ΔGrxn = -50199 J + (8.314 J/(mol K) * 298.15 K * ln(0.0062))
= -50199 J + (2475.97 J)
= -47723.03 J