balance the following redox reactions in acid...
balance the following redox reactions in acidic and alkaline solutions, then answer the stoichiometric problems based on the chemical reaction. no points for no solutions.\nbro₃⁻ + fe²⁺ → br⁻ + fe³⁺ (acidic solution) 20 pts\na. how many moles of fe²⁺ are required to react completely with 1 mole of bro₃⁻?\nb. if 0.5 moles of bro₃⁻ react, how many moles of fe³⁺ are produced?\nc. how many moles of water are formed when 2 moles of bro₃⁻ react?\nbr⁻ + mno₄⁻ → mno₂ + bro₃⁻ (alkaline solution) 20 pts\na. how many moles of mno₄⁻ are required to react completely with 4 moles of br⁻?\nb. if 3 moles of br⁻ react, how many moles of oh⁻ are produced?\nc. how many moles of mno₂ form when 5 moles of mno₄⁻ react?
Answer
# Explanation:
## Step1: Balance the acidic - solution reaction
For the reaction $\mathrm{BrO}_{3}^{-}+\mathrm{Fe}^{2 + }\to\mathrm{Br}^{-}+\mathrm{Fe}^{3 + }$ in acidic solution:
1. First, identify the oxidation and reduction half - reactions.
- Oxidation half - reaction: $\mathrm{Fe}^{2 + }\to\mathrm{Fe}^{3 + }+e^{-}$
- Reduction half - reaction: $\mathrm{BrO}_{3}^{-}+6H^{+}+6e^{-}\to\mathrm{Br}^{-}+3H_{2}O$
2. Multiply the oxidation half - reaction by 6 to balance the electrons:
- $6\mathrm{Fe}^{2 + }\to6\mathrm{Fe}^{3 + }+6e^{-}$
- $\mathrm{BrO}_{3}^{-}+6H^{+}+6e^{-}\to\mathrm{Br}^{-}+3H_{2}O$
- The balanced reaction is $\mathrm{BrO}_{3}^{-}+6\mathrm{Fe}^{2 + }+6H^{+}=\mathrm{Br}^{-}+6\mathrm{Fe}^{3 + }+3H_{2}O$
## Step2: Answer part (a) of the acidic - solution reaction
From the balanced equation $\mathrm{BrO}_{3}^{-}+6\mathrm{Fe}^{2 + }+6H^{+}=\mathrm{Br}^{-}+6\mathrm{Fe}^{3 + }+3H_{2}O$, the mole ratio of $\mathrm{BrO}_{3}^{-}$ to $\mathrm{Fe}^{2 + }$ is $1:6$. So, if we have 1 mole of $\mathrm{BrO}_{3}^{-}$, the number of moles of $\mathrm{Fe}^{2 + }$ required is 6 moles.
## Step3: Answer part (b) of the acidic - solution reaction
The mole ratio of $\mathrm{BrO}_{3}^{-}$ to $\mathrm{Fe}^{3 + }$ is $1:6$. If 0.5 moles of $\mathrm{BrO}_{3}^{-}$ react, the number of moles of $\mathrm{Fe}^{3 + }$ produced is $n = 0.5\times6=3$ moles.
## Step4: Answer part (c) of the acidic - solution reaction
The mole ratio of $\mathrm{BrO}_{3}^{-}$ to $\mathrm{H}_{2}O$ is $1:3$. If 2 moles of $\mathrm{BrO}_{3}^{-}$ react, the number of moles of $\mathrm{H}_{2}O$ formed is $n = 2\times3 = 6$ moles.
## Step5: Balance the alkaline - solution reaction
For the reaction $\mathrm{Br}^{-}+\mathrm{MnO}_{4}^{-}\to\mathrm{MnO}_{2}+\mathrm{BrO}_{3}^{-}$ in alkaline solution:
1. Identify the oxidation and reduction half - reactions.
- Oxidation half - reaction: $\mathrm{Br}^{-}+6\mathrm{OH}^{-}\to\mathrm{BrO}_{3}^{-}+3H_{2}O + 6e^{-}$
- Reduction half - reaction: $\mathrm{MnO}_{4}^{-}+2H_{2}O+3e^{-}\to\mathrm{MnO}_{2}+4\mathrm{OH}^{-}$
2. Multiply the reduction half - reaction by 2 to balance the electrons:
- $\mathrm{Br}^{-}+6\mathrm{OH}^{-}\to\mathrm{BrO}_{3}^{-}+3H_{2}O + 6e^{-}$
- $2\mathrm{MnO}_{4}^{-}+4H_{2}O+6e^{-}\to2\mathrm{MnO}_{2}+8\mathrm{OH}^{-}$
- The balanced reaction is $\mathrm{Br}^{-}+2\mathrm{MnO}_{4}^{-}+H_{2}O=\mathrm{BrO}_{3}^{-}+2\mathrm{MnO}_{2}+2\mathrm{OH}^{-}$
## Step6: Answer part (a) of the alkaline - solution reaction
From the balanced equation $\mathrm{Br}^{-}+2\mathrm{MnO}_{4}^{-}+H_{2}O=\mathrm{BrO}_{3}^{-}+2\mathrm{MnO}_{2}+2\mathrm{OH}^{-}$, the mole ratio of $\mathrm{Br}^{-}$ to $\mathrm{MnO}_{4}^{-}$ is $1:2$. So, if we have 4 moles of $\mathrm{Br}^{-}$, the number of moles of $\mathrm{MnO}_{4}^{-}$ required is 8 moles.
## Step7: Answer part (b) of the alkaline - solution reaction
The mole ratio of $\mathrm{Br}^{-}$ to $\mathrm{OH}^{-}$ is $1:2$. If 3 moles of $\mathrm{Br}^{-}$ react, the number of moles of $\mathrm{OH}^{-}$ produced is $n = 3\times2=6$ moles.
## Step8: Answer part (c) of the alkaline - solution reaction
The mole ratio of $\mathrm{MnO}_{4}^{-}$ to $\mathrm{MnO}_{2}$ is $1:1$. If 5 moles of $\mathrm{MnO}_{4}^{-}$ react, the number of moles of $\mathrm{MnO}_{2}$ formed is 5 moles.
# Answer:
a. (Acidic solution) 6 moles
b. (Acidic solution) 3 moles
c. (Acidic solution) 6 moles
a. (Alkaline solution) 8 moles
b. (Alkaline solution) 6 moles
c. (Alkaline solution) 5 moles