a.) which of the following reactions should have the larger emf under standard conditions & why?
CuSO4 (aq) + Pb (s) goes to PbSO4 (s) + Cu (s);
Cu(NO3)2 (aq) + Pb(s) goes to Pb(NO3)2(aq) + Cu(s)
b.) Calculate delta G for the reaction.
a.) To determine which reaction should have the larger electromotive force (emf) under standard conditions, you need to consider the standard reduction potentials of the elements involved.
The emf of a cell is given by the difference in reduction potentials between the two half-reactions. The higher the reduction potential, the greater the tendency for reduction to occur. Therefore, the reaction with the larger emf will be the one with a greater difference in reduction potentials between the two half-reactions.
For reaction (CuSO4 (aq) + Pb (s) → PbSO4 (s) + Cu (s)), the half-reactions are:
Cu2+ + 2e- → Cu (E° = +0.34 V) [from table]
Pb2+ + 2e- → Pb (E° = -0.126 V) [from table]
The reduction potential for the Cu half-reaction is higher than that of the Pb half-reaction. Thus, the Cu half-reaction is more likely to occur as a reduction half-reaction, making the overall reaction spontaneous.
For reaction (Cu(NO3)2 (aq) + Pb(s) → Pb(NO3)2(aq) + Cu(s)), the half-reactions are:
Cu2+ + 2e- → Cu (E° = +0.34 V) [from table]
Pb2+ + 2e- → Pb (E° = -0.126 V) [from table]
The reduction potential for the Cu half-reaction is also higher than that of the Pb half-reaction. Therefore, the reaction will be spontaneous.
In both cases, the Cu half-reaction has the higher reduction potential. Therefore, both reactions should have the same emf under standard conditions.
b.) To calculate ΔG (the change in Gibbs free energy) for a reaction, you can use the equation:
ΔG = -nFE
where ΔG is the change in Gibbs free energy, n is the number of moles of electrons transferred, F is Faraday's constant (approximately 96,485 C/mol), and E is the cell potential.
From the standard reduction potentials obtained in part (a), we see that the Cu half-reaction is the reduction half-reaction in both reactions. Therefore, n = 2 (since each copper ion is reduced by the gain of 2 electrons).
For both reactions, the cell potential (E) is the same, as the reactions have the same emf. The value of E can be calculated by subtracting the reduction potential of the anode (where oxidation occurs) from the reduction potential of the cathode (where reduction occurs). E = E(cathode) - E(anode).
Once you have calculated the value of E, you can proceed to calculate ΔG using the given equation.
Please provide the standard reduction potentials of the half-reactions involved in order to calculate ΔG for the desired reaction.
a) To determine which reaction should have the larger emf (electromotive force) under standard conditions, we can look at the standard reduction potentials of the half-reactions involved.
The standard reduction potential for the Cu2+(aq) + 2e- → Cu(s) half-reaction is +0.34 V.
The standard reduction potential for the Pb2+(aq) + 2e- → Pb(s) half-reaction is -0.13 V.
In the first reaction, Cu2+ is reduced to Cu, and Pb is oxidized to Pb2+. The total emf of the cell can be calculated by subtracting the reduction potential of the oxidation reaction from the reduction potential of the reduction reaction.
For the first reaction, the emf would be: E = Ered(Cu2+/Cu) - Ered(Pb2+/Pb)
E = +0.34 V - (-0.13 V) = +0.47 V
In the second reaction, Cu2+ is reduced to Cu, and Pb is oxidized to Pb2+. The total emf of the cell can be calculated in the same way.
For the second reaction, the emf would be: E = Ered(Cu2+/Cu) - Ered(Pb2+/Pb)
E = +0.34 V - (-0.13 V) = +0.47 V
Therefore, both reactions should have the same emf under standard conditions (+0.47 V) since they involve the same half-reactions.
b) To calculate the Delta G (change in Gibbs free energy) for a reaction, we can use the equation:
Delta G = -nFE
Where:
Delta G is the change in Gibbs free energy (in Joules)
n is the number of moles of electrons transferred in the reaction
F is Faraday's constant (96,485 C/mol)
E is the cell emf (in volts)
Since we know the emf for the reaction is 0.47 V, and the number of moles of electrons transferred in this reaction is 2 electrons (since 2 electrons are involved in the reduction of each Cu2+ ion), we can substitute these values into the equation.
Delta G = - (2 mol)(96,485 C/mol)(0.47 V)
Delta G = -91,451 J
Therefore, the Delta G for the reaction is -91,451 J.
In #1, formation of Pb to Pb^+ should be smaller than Pb ==> PbSO4 because the formation of PbSO4 drives the reaction to the right.
#2. dG = -nFEocell