Calculate ΔGo (in J) for 1 moles of electron(s) in each half-reaction and a standard potential of 0.34 V.
I got .86
Calculate Keq for 2 moles of electron(s) in each half-reaction and a standard potential of (-0.44) V.
I got -.36
Calculate the potential for: Br2(l) + 2Cl-(aq) → Cl2(g) + 2Br-(aq) Use Eº = -.300 V
when the concentrations of the soluble species are as follows:
[Cl-] = 0.46 M, [Br-] = 2.0x10-3 M, PCl2 = 3.9x10-2 atm
I got -.32
All of these are wrong :( can someone please help?
To calculate ΔGo (standard Gibbs free energy change) for each half-reaction, you need to use the formula:
ΔGo = -nFΔEo
where:
- ΔGo is the standard Gibbs free energy change in Joules (J)
- n is the number of moles of electron(s) involved in the half-reaction
- F is the Faraday constant (which is approximately 96485 C/mol)
- ΔEo is the standard potential in Volts (V)
Let's calculate each one:
1) For 1 mole of electron(s) and a standard potential of 0.34 V:
ΔGo = -1 * 96485 C/mol * 0.34 V
ΔGo = -32840 J
So, for this case, ΔGo is -32840 J.
2) For 2 moles of electron(s) and a standard potential of -0.44 V:
ΔGo = -2 * 96485 C/mol * -0.44 V
ΔGo = -84505 J
So, for this case, ΔGo is -84505 J.
Now, let's move on to calculating Keq:
The expression for the equilibrium constant Keq can be expressed as:
Keq = e^(ΔGo / (RT))
where:
- Keq is the equilibrium constant
- ΔGo is the standard Gibbs free energy change
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (K)
3) For 2 moles of electron(s) and a standard potential of -0.44 V:
ΔGo = -84505 J (from the previous calculation)
Let's assume a temperature of, for example, 298 K:
Keq = e^(-84505 J / (8.314 J/(mol·K) * 298 K))
Keq = 1.56
So, for this case, Keq is approximately 1.56.
Lastly, for the given reaction and concentrations, we can use the Nernst equation to calculate the potential:
E = Eº - (0.0592 V / z) * log(Q)
where:
- E is the cell potential
- Eº is the standard potential
- z is the number of electrons transferred in the reaction
- Q is the reaction quotient, calculated using the concentrations or pressures of the species involved
4) For the reaction: Br2(l) + 2Cl-(aq) → Cl2(g) + 2Br-(aq)
Considering Eº = -0.300 V:
E = -0.300 V - (0.0592 V / 2) * log([Cl-]^2 * [Br-]^2 / [Br2] * [PCl2])
E = -0.300 V - (0.0296 V) * log((0.46 M)^2 * (2.0x10-3 M)^2 / (3.9x10-2 atm) * (1 atm))
E = -0.32 V
So, for this case, the potential is approximately -0.32 V.
Please note that the calculations were made assuming standard conditions unless specified otherwise, and make sure to double-check the units and calculations to avoid mistakes.