The electrochemical cell described by the balanced chemical equation has a standard emf of -0.25 V. Calculate the value (kJ) for the ΔG˚ of the cell. Round your answer to 3 significant figures.
MnO2(s) + Pb2+(aq) → Mn2+(aq) + PbO2(s)
St. Red. Pot. (V)
PbO2/Pb2+ +1.46
MnO2/Mn2+ +1.21
Faraday's Constant
F = 96485 C
DGo = -nEoF
Substitute and solve for DGo
To calculate the value for the standard Gibbs free energy change (ΔG˚) of the electrochemical cell, you can use the equation:
ΔG˚ = -nFΔE˚
Where:
- ΔG˚ is the standard Gibbs free energy change of the cell,
- n is the number of moles of electrons transferred in the balanced equation,
- F is Faraday's constant (96485 C), and
- ΔE˚ is the standard electromotive force (emf) of the cell.
First, let's determine the number of moles of electrons transferred (n) in the balanced equation:
MnO2(s) + Pb2+(aq) → Mn2+(aq) + PbO2(s)
From the balanced equation, it can be observed that the electrons are not present on both sides of the equation. To balance the charge, 2 electrons are transferred.
Now, let's substitute the values into the equation:
n = 2 (from the balanced equation)
F = 96485 C (Faraday's constant)
ΔE˚ = -0.25 V (standard emf of the electrochemical cell)
ΔG˚ = -nFΔE˚
ΔG˚ = -(2)(96485 C)(-0.25 V)
ΔG˚ = 48342.5 C V
To convert from Coulomb Volts to kilojoules (kJ), you need to use the equation:
1 C V = 1 J
1 kJ = 1000 J
Therefore:
ΔG˚ = (48342.5 C V) / (1000 J/kJ)
ΔG˚ ≈ 48.3 kJ
So, the value for the ΔG˚ of the cell is approximately 48.3 kJ.