Here we consider instead dissolving guns (which we will assume are pure iron) with sulfuric acid.
Fe+H2SO4->FeSO4+H2
It is apparent that they must use an outside source of electrical power to drive the dissolution of the iron. They use a small current so that the dissolution proceeds with the minimum voltage required. Assume standard values for the electrochemical potentials.
How much electrical energy supplied this way is thus required to dissolve an additional 1 kg of iron? Give your answer in kJ.
To determine the amount of electrical energy required to dissolve 1 kg of iron, we need to calculate the number of moles of iron being dissolved and then use the equation E = nFΔE, where E is the electrical energy, n is the number of moles, F is the Faraday constant, and ΔE is the change in potential.
First, we need to calculate the number of moles of iron in 1 kg. The molar mass of iron (Fe) is approximately 55.845 g/mol. Therefore, the number of moles can be calculated as follows:
Number of moles of iron = mass of iron / molar mass of iron
Number of moles of iron = 1000 g / 55.845 g/mol = 17.91 mol
Next, we need to consider the stoichiometry of the reaction. From the balanced equation, it is clear that one mole of iron reacts with one mole of sulfuric acid to produce one mole of iron sulfate. Therefore, the number of moles of iron dissolved is equal to the number of moles of sulfuric acid consumed.
Since iron is being oxidized during this reaction, the change in potential (ΔE) can be determined using the standard reduction potential (E°) for the iron-sulfuric acid half-reaction:
Fe(s) → Fe2+(aq) + 2e− (E° = -0.44 V)
The negative sign is used because the standard reduction potential represents the reduction process rather than the oxidation process. The change in potential (ΔE) in this case would be -0.44 V.
The Faraday constant (F) is approximately equal to 96,485 C/mol.
Now we can calculate the electrical energy (E):
E = nFΔE
E = 17.91 mol * 96,485 C/mol * -0.44 V
Using the given values and performing the calculation will yield the amount of electrical energy required to dissolve an additional 1 kg of iron in kJ.