35 lezione

It seems you have a series of chemical equations and balancing problems. Let's go through them step by step to ensure they are correctly balanced, especially focusing on the principles of oxidation-reduction (redox) reactions. ### Problem 1: **Equation:** \[ \text{Cl}_2 + \text{P} \rightarrow \text{PCl}_3 \] #### Step-by-Step Solution: 1. **Identify Oxidation States:** - Cl in \(\text{Cl}_2\) is \(0\). - P in \(\text{P}\) is \(0\). - P in \(\text{PCl}_3\) is +3. - Cl in \(\text{PCl}_3\) is -3. 2. **Determine Oxidation and Reduction:** - Phosphorus (P) changes from 0 to +3, so it is oxidized. - Chlorine (Cl) changes from 0 to -3, so it is reduced. 3. **Balance the Equation:** - Start with the unbalanced equation: \[ \text{Cl}_2 + \text{P} \rightarrow \text{PCl}_3 \] - Balance P and Cl separately. - For every 1 atom of P, we need 3 atoms of Cl to form \(\text{PCl}_3\): \[ \text{Cl}_2 + \text{P} \rightarrow \text{PCl}_3 \] 4. **Balance the Chlorine:** - To balance chlorine, add a coefficient of 3 in front of \(\text{Cl}_2\): \[ 3\text{Cl}_2 + \text{P} \rightarrow \text{PCl}_3 \] - Now, we have 6 Cl atoms on the left and 3 Cl atoms on the right. To balance this, add a coefficient of 2 in front of \(\text{PCl}_3\): \[ 3\text{Cl}_2 + \text{P} \rightarrow 2\text{PCl}_3 \] 5. **Verify Oxidation States:** - Chlorine is reduced from \(0\) to \(-3\). - Phosphorus is oxidized from \(0\) to \(+3\). 6. **Final Balanced Equation:** \[ 3\text{Cl}_2 + \text{P} \rightarrow 2\text{PCl}_3 \] ### Problem 2: **Equation:** \[ \text{Fe} + \text{S} \rightarrow \text{Fe}_2\text{S}_3 \] #### Step-by-Step Solution: 1. **Identify Oxidation States:** - Fe in \(\text{Fe}\) is \(0\). - S in \(\text{S}\) is \(0\). - Fe in \(\text{Fe}_2\text{S}_3\) is +2.5. - S in \(\text{Fe}_2\text{S}_3\) is -1.5. 2. **Determine Oxidation and Reduction:** - Iron (Fe) changes from 0 to +2.5, so it is oxidized. - Sulfur (S) changes from 0 to -1.5, so it is reduced. 3. **Balance the Equation:** - Start with the unbalanced equation: \[ \text{Fe} + \text{S} \rightarrow \text{Fe}_2\text{S}_3 \] - Balance Fe and S s