Question

When the following skeletal equation is balanced under basic conditions, what are the coefficients of the species shown?

SO₄^2-+  F^- = F₂+  SO₃^2-

Water appears in the balanced equation as a  (reactant, product, neither) with a coefficient of............ (Enter 0 for neither.)

Which species is the reducing agent?

Answer 1

Answer 2

Given Skeletal Equation:

$$S{O}_4^{2-}+F^{-}\to F_2+S{O}_3^{2-}$$

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Step 1: Assign Oxidation States

• Sulfur (S):

• In $S{O}_4^{2-}$: $+6$ (each O is $-2$, total charge $-8+S=-2⟹S=+6$).

• In $S{O}_3^{2-}$: $+4$ (each O is $-2$, total $-6+S=-2⟹S=+4$).

• Fluorine (F):

• In $F^{-}$: $-1$.

• In $F_2$: $0$ (elemental form).

• 
  

Key Observations:

• Sulfur is reduced ($+6\to +4$).

• Fluorine is oxidized ($-1\to 0$).

• 
  

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Step 2: Write Half-Reactions

1. Reduction Half-Reaction (Sulfur):

   $$S{O}_4^{2-}\to S{O}_3^{2-}$$

• Balance O by adding $H_{2}O$:

  $$S{O}_4^{2-}\to S{O}_3^{2-}+H_{2}O$$

• Balance H by adding $O{H}^{-}$ (basic conditions):

  $$S{O}_4^{2-}+H_{2}O\to S{O}_3^{2-}+2O{H}^{-}$$

• Balance charge with electrons:

  $$S{O}_4^{2-}+H_{2}O+2e^{-}\to S{O}_3^{2-}+2O{H}^{-}$$

2. Oxidation Half-Reaction (Fluorine):

   $$F^{-}\to F_2$$

• Balance F atoms:

  $$2F^{-}\to F_2$$

• Balance charge with electrons:

  $$2F^{-}\to F_2+2e^{-}$$

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Step 3: Combine Half-Reactions

• Both half-reactions involve 2 electrons. Simply add them:

  $$S{O}_4^{2-}+H_{2}O+2F^{-}\to S{O}_3^{2-}+2O{H}^{-}+F_2$$

Balanced Equation:

$$\boxed{{\displaystyle 1}}S{O}_4^{2-}+\boxed{{\displaystyle 2}}{F}^{-}\to \boxed{{\displaystyle 1}}{F}_2+\boxed{{\displaystyle 1}}S{O}_3^{2-}+\boxed{{\displaystyle 2}}O{H}^{-}$$

• Water ($H_{2}O$): Appears as a reactant with a coefficient of $\boxed{{\displaystyle 1}}$.

• 
  

• Reducing Agent: Species that is oxidized ($F^{-}$), so $\boxed{{\displaystyle F^{-}}}$.

• 
  

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Answer:

1. Coefficients:

• $S{O}_4^{2-}$: 1

• $F^{-}$: 2

• $F_2$: 1

• $S{O}_3^{2-}$: 1

• $O{H}^{-}$: 2

2. Water: Reactant with coefficient $\boxed{{\displaystyle 1}}$.

3. Reducing Agent: $\boxed{{\displaystyle F^{-}}}$.

4. 
   

Balanced Equation:

$$\boxed{{\displaystyle 1}}S{O}_4^{2-}+\boxed{{\displaystyle 2}}{F}^{-}\to \boxed{{\displaystyle 1}}{F}_2+\boxed{{\displaystyle 1}}S{O}_3^{2-}+\boxed{{\displaystyle 2}}O{H}^{-}$$