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Question 19.3: Balancing Redox Reactions Occurring in Basic Solution Balanc...

Balancing Redox Reactions Occurring in Basic Solution

Balance the equation occurring in basic solution:

I^{-}(a q)+ MnO _4  ^{-}(a q) \longrightarrow I_{2}(a q)+ MnO _2(S)

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To balance redox reactions occurring in basic solution, follow the half-reaction method outlined in Examples 19.1 and 19.2, but add an extra step to neutralize the acid with OH^{-} as shown in step 3.

1. Assign oxidation states.
2. Separate the overall reaction into two half-reactions. Oxidation: I^{-}(a q) \longrightarrow I_{2}(a q)

Reduction: MnO _4  ^{-}(a q) \longrightarrow MnO _2(S)
3. Balance each half-reaction with respect to mass.

  •  Balance all elements other than H and O.
  •  Balance O by adding H_{2}O.
  •  Balance H by adding H^{+}.
  • Neutralize H^{+} by adding enough OH^{-} to neutralize each H^{+}. Add the same number of OH^{-} ions to each side of the equation.
 \left\{\begin{array}{l} 2  I ^{-}(a q) \longrightarrow I _2(a q) \\ MnO _4{ }^{-}(a q) \longrightarrow MnO _2(s) \end{array}\right.

\left\{\begin{array}{l} 2  I ^{-}(a q) \longrightarrow I _2(a q) \\ MnO _4{ }^{-}(a q) \longrightarrow MnO _2(s) +2  H_{2}O (l)\end{array}\right.

 

\left\{\begin{array}{l} 2  I ^{-}(a q) \longrightarrow I _2(a q) \\ 4  H^{+}(aq)+MnO _4{ }^{-}(a q) \longrightarrow MnO _2(s) +2  H_{2}O (l)\end{array}\right.

 

\left\{\begin{array}{l} 2  I ^{-}(a q) \longrightarrow I _2(a q) \\ \underbrace{4  H^{+}(aq)+4  OH^{-}(aq) }_{4  H_{2}O (l) }+MnO _4{ }^{-}(a q) \longrightarrow MnO _2(s) +2  H_{2}O (l) + 4  OH^{-}(aq)\end{array}\right.

4. Balance each half-reaction with respect to charge. 2  I ^{-}(a q) \longrightarrow I _2(a q) +2  e^{-}

 

4  H_{2}O (l) +MnO _4{ }^{-}(a q)+3  e^{-} \longrightarrow MnO _2(s) +2  H_{2}O (l) + 4  OH^{-}(aq)
5. Make the number of electrons in both half-reactions equal. 3[2 I ^{-}(a q) \longrightarrow I _2(a q) +2  e^{-}]

 

6  I ^{-}(a q) \longrightarrow 3  I _2(a q) +6  e^{-}

 

2[4  H_{2}O (l) +MnO _4{ }^{-}(a q)+3  e^{-} \longrightarrow MnO _2(s) +2  H_{2}O (l) + 4  OH^{-}(aq)]

 

8  H_{2}O (l) +2  MnO _4{ }^{-}(a q)+6  e^{-} \longrightarrow 2  MnO _2(s) +4  H_{2}O (l) +8  OH^{-}(aq)
6. Add the half-reactions together. 6  I ^{-}(a q) \longrightarrow 3  I _2(a q) +\cancel{6  e^{-}}

4  \cancel{8} H_{2}O (l) +2  MnO _4{ }^{-}(a q)+\cancel{6  e^{-}} \longrightarrow2  MnO _2(s) +\cancel{4  H_{2}O (l)}+8  OH^{-}(aq) \\ \overline{6  I ^{-}(a q)+ 4  H_{2}O (l) +2  MnO _4{ }^{-}(a q) \longrightarrow 3  I _2(a q)+2  MnO _2(s) +8  OH^{-}(aq)}
7. Verify that the reaction is balanced.
Reactants Products
6 I 6 I
8 H 8 H
2 Mn 2 Mn
12 O 12 O
 -8 charge -8 charge

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