The standard electrode potentials for zn/zn and cu/cu are -0.76v and
0.34v respectively at 298k.calculate the free energy change when a zinc rod is dip into a zinc tetraoxosulphate(vi) solution and a copper rod is dip into a copper(ii)tetraoxosulphate(vi) solution, both are connected together by a salt bridge
Well, let's dip into this electrochemical circus, shall we? 🎪
To calculate the free energy change (ΔG), we can use the equation:
ΔG = -nFE
Where:
ΔG is the free energy change,
n is the number of electrons transferred,
F is Faraday's constant (96,485 C/mol),
and E is the cell potential.
In this case, for the zinc half-cell (Zn/Zn2+), the electrode potential is -0.76V, and for the copper half-cell (Cu2+/Cu), the electrode potential is 0.34V.
Now, it's important to note that the cell potential (Ecell) is the difference between the two half-cell potentials:
Ecell = E(cathode) - E(anode)
In this scenario, copper is the cathode (+0.34V) and zinc is the anode (-0.76V), so the cell potential is:
Ecell = 0.34V - (-0.76V) = 1.1V
Since the half-reaction involves the transfer of 2 electrons (Zn → Zn2+ + 2e-), we can plug these values into our trusty equation:
ΔG = -nFE
= -(2)(96,485 C/mol)(1.1V)
= -211,267 J/mol
Voila! The free energy change (ΔG) is -211,267 J/mol. Keep in mind that this answer assumes a standard state and ideal conditions. And as we all know, the circus isn't always so straightforward. ðŸŽ
To calculate the free energy change, ΔG, for the given electrochemical cell, we can use the Nernst equation:
ΔG = -nFE
Where:
ΔG = Free energy change
n = Number of moles of electrons transferred
F = Faraday's constant (96,485 C/mol)
E = Cell potential
First, let's determine the overall reaction and the number of moles of electrons transferred.
The cell reaction is as follows:
Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)
In this reaction, one mole of electrons is transferred, as the coefficient of Cu2+ is 1.
Now, we need to find the cell potential, E. The cell potential is calculated by subtracting the standard electrode potential of the anode (Zn/Zn2+) from the standard electrode potential of the cathode (Cu2+/Cu).
E = E(cathode) - E(anode)
E = 0.34 V - (-0.76 V)
E = 1.10 V
Now, we can calculate the free energy change using the Nernst equation.
ΔG = -nFE
ΔG = -(1 mol × 96,485 C/mol × 1.10 V)
ΔG = -106,133.5 J
The negative sign indicates that the reaction is spontaneous, and the magnitude of ΔG represents the amount of work that can be extracted from the cell.
Therefore, the free energy change when a zinc rod is dipped into a zinc tetraoxosulphate(vi) solution and a copper rod is dipped into a copper(ii)tetraoxosulphate(vi) solution, both connected by a salt bridge, is approximately -106,133.5 J.
To calculate the free energy change (ΔG) when a zinc rod is dipped into a zinc tetraoxosulphate(VI) solution and a copper rod is dipped into a copper(II) tetraoxosulphate(VI) solution connected by a salt bridge, we need to use the Nernst equation.
The Nernst equation relates the standard electrode potential (E°) of a half-cell to the actual electrode potential (E) under non-standard conditions and the concentration of the electroactive species. It is given by:
E = E° - (RT/nF) * ln(Q)
Where:
E = Actual electrode potential (V)
E° = Standard electrode potential (V)
R = Gas constant (8.314 J/mol·K)
T = Temperature in Kelvin (298 K in this case)
n = Number of electrons involved in the redox reaction
F = Faraday's constant (96485 C/mol)
Q = Reaction quotient (ratio of product concentrations to reactant concentrations)
Let's calculate the ΔG for each half-cell separately.
For the zinc half-cell (Zn/Zn2+):
E° = -0.76 V
Assuming the concentration of Zn2+ is 1M, the reaction quotient is given by:
Q = [Zn2+]/[Zn] = 1/1 = 1
Substituting the values into the Nernst equation:
E = -0.76 V - (8.314 J/mol·K * 298 K / (2 * 96485 C/mol)) * ln(1)
E = -0.76 V
Since we are assuming standard conditions, the actual electrode potential is equal to the standard electrode potential.
For the copper half-cell (Cu/Cu2+):
E° = 0.34 V
Assuming the concentration of Cu2+ is also 1M, the reaction quotient is given by:
Q = [Cu2+]/[Cu] = 1/1 = 1
Substituting the values into the Nernst equation:
E = 0.34 V - (8.314 J/mol·K * 298 K / (2 * 96485 C/mol)) * ln(1)
E = 0.34 V
Again, the actual electrode potential is equal to the standard electrode potential.
Since the actual potentials for both half-cells are equal to their standard potentials, it indicates that no spontaneous redox reaction will take place. Therefore, there will be no change in free energy (ΔG = 0) when the zinc and copper electrodes are connected in this configuration.