Asked by Brianna
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.
2.Calculate the following values at 25 degrees celcius. Eocell, Ecell, deltaGo rxn, and delta Grxn
Answers
Answered by
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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
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