Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:
In this case, the cathode and anode both are same. So, is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
= 0.0222 M
= 2.22 M
Now put all the given values in the above equation, we get:
Therefore, the cell emf for this cell is 0.118 V
Answer:
Explanation:
The chemical expression for the reaction between iron and oxygen is:
4Fe(s) + 3O₂ (g) 2Fe₂O₃ (s)
The number of moles of Fe = mass of Fe/ molecular mass of Fe
The number of moles of Fe = 56 g/ 55.845 g/mol
The number of moles of Fe = 1.002 moles of Fe
The number of moles of oxygen = mass of oxygen/ molecular mass of oxygen
The number of moles of oxygen = 156 g /32 g/mol
The number of moles of oxygen = 4.875 moles of oxygen
Assume that Fe is the limiting reactant, the number of Fe₂O₃ can be calculated as:
moles of Fe₂O₃ = 1.002 mole of Fe × 2 moles of Fe₂O₃/ 4 moles of Fe
moles of Fe₂O₃ = 0.501 mole of Fe₂O₃
Assume that O₂ is the limiting factor, the number of Fe₂O₃ is:
moles of Fe₂O₃ = 4.875 moles of O₂ × 2 moles of Fe₂O₃/ 3 moles of O₂
moles of Fe₂O₃ = 3.25 mole of Fe₂O₃
Thus, after the reaction is complete, Fe and O₂ contain different moles of Fe₂O₃. Only Fe gets consumed in the reaction and it is the limiting factor.
Answer: pH = 1.1
Explanation: see attachment
