Free-energy change, ΔG∘, is related to cell potential, E∘, by the equation
ΔG∘=−nFE∘
where n is the number of moles of electrons transferred and F=96,500C/(mol e−) is the Faraday constant. When E∘ is measured in volts, ΔG∘ must be in joules since 1 J=1 C⋅V.
Calculate the standard free-energy change at 25 ∘C for the following reaction:
Mg(s)+Fe2+(aq)→Mg2+(aq)+Fe(s)
Express your answer to three significant figures and include the appropriate units.
I thought it was 373 kj*mol^-1 buts its saying something is wrong
If you will show what you did perhaps we can find the problem. Other than guessing, however, we don't know what you've done.
To calculate the standard free-energy change (ΔG°) for the given reaction, we need to use the equation ΔG° = -nFE°, where n is the number of moles of electrons transferred and F is the Faraday constant.
First, let's determine the values for n and E°:
In the given reaction, one mole of electrons is transferred. Therefore, n = 1.
To find the standard cell potential E°, we can use the standard reduction potentials for each half-reaction involved in the reaction:
Mg2+(aq) + 2e- → Mg(s): E° = -2.37 V
Fe2+(aq) + 2e- → Fe(s): E° = -0.44 V
Since the standard cell potential (E°) is the difference between these two reduction potentials, we have:
E° = E°(Mg2+(aq) + 2e- → Mg(s)) - E°(Fe2+(aq) + 2e- → Fe(s))
E° = -2.37 V - (-0.44 V)
E° = -1.93 V
Now we can calculate ΔG° using the equation:
ΔG° = -nFE°
Substituting the values, we have:
ΔG° = -(1 mol)(96,500 C/mol e-)(-1.93 V)
ΔG° = 186,845 J
Finally, we can convert the result from joules to kilojoules:
ΔG° = 186,845 J = 186.85 kJ
Therefore, the standard free-energy change at 25 °C for the given reaction is approximately -186.85 kJ (or 187 kJ) in units of kilojoules.
To calculate the standard free-energy change (ΔG∘) at 25°C for a given reaction, we can use the equation ΔG∘=−nFE∘. In this equation, n represents the number of moles of electrons transferred, F is the Faraday constant (96,500 C/mol e−), and E∘ is the cell potential.
Let's break down the problem:
1. Determine the number of moles of electrons transferred (n):
In the given reaction: Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s), we see that Mg loses two electrons while Fe2+ gains two electrons. Therefore, the number of moles of electrons transferred (n) is 2.
2. Find the cell potential (E∘):
To determine the cell potential, we need to refer to a table of standard reduction potentials. The standard reduction potential for the reaction Fe2+(aq) + 2e− → Fe(s) is -0.44 V. The negative sign indicates that this reaction is a reduction.
3. Calculate the standard free-energy change (ΔG∘):
Using the given values, we can substitute n = 2 and E∘ = -0.44 V into the equation ΔG∘=−nFE∘ to find ΔG∘:
ΔG∘ = -(2)(96,500 C/mol e−)(-0.44 V)
ΔG∘ = 85,120 J/mol
4. Convert the units:
Since the question asks for the answer in joules, we don't need to convert the units.
Therefore, the standard free-energy change at 25°C for the given reaction is 85,120 J/mol.