The galvanic cell described by the cell notation has a standard emf (electromotive force) of 4.05 V. Calculate the value (kJ) for the ΔG˚ of the cell. Round your answer to 3 significant figures.
Mn(s) l Mn2+(aq) ll F2(g) l F-(aq) l Pt(s)
St. Red. Pot. (V)
Mn2+/Mn -1.18
F2/F- +2.87
Faraday's Constant
F = 96485 C
To calculate the value of ΔG° (standard Gibbs free energy change) of the galvanic cell, we can use the formula:
ΔG° = -nFE°
Where:
ΔG° is the standard Gibbs free energy change (in kJ)
n is the number of moles of electrons transferred
F is Faraday's constant (96485 C)
E° is the standard emf (electromotive force) of the cell (in V)
First, let's determine the value of n, the number of moles of electrons transferred. Looking at the balanced equation, we can see that in the reduction half-reaction:
Mn2+(aq) + 2e- → Mn(s)
One mole of Mn2+ ions requires two moles of electrons. So n = 2 (2 moles of electrons).
Next, we substitute the given values into the formula:
ΔG° = -nFE°
= -(2)(96485 C)(4.05 V)
Calculating this expression, we get:
ΔG° = -783,759 C·V
To convert this value from coulombs·volts to kilojoules, we use the conversion factor:
1 C·V = 1 J (Joule)
1 kJ = 1000 J
Therefore, we have:
ΔG° = -783,759 J
Converting J to kJ:
ΔG° = -783.759 kJ
Finally, rounding the answer to three significant figures:
ΔG° ≈ -784 kJ
To calculate the value (kJ) for the ΔG˚ of the cell, we can use the formula:
ΔG˚ = -nFE˚
where:
- ΔG˚ is the change in Gibbs free energy (in kJ),
- n is the number of moles of electrons transferred during the cell reaction,
- F is Faraday's constant (96485 C/mol),
- and E˚ is the standard emf of the cell (in V).
First, we need to identify the number of electrons transferred during the cell reaction. Looking at the cell notation:
Mn(s) l Mn2+(aq) ll F2(g) l F-(aq) l Pt(s)
We can see that Mn loses 2 electrons to form Mn2+ and F2 gains 2 electrons to form F-. Therefore, the number of moles of electrons transferred (n) is 2.
Now we can substitute the values into the formula:
ΔG˚ = -nFE˚
ΔG˚ = -(2 mol)(96485 C/mol)(4.05 V)
Calculating this:
ΔG˚ = -196393 kJ
Rounded to 3 significant figures:
ΔG˚ ≈ -196000 kJ
Therefore, the value for the ΔG˚ of the cell is approximately -196000 kJ.