Certainly! To write balanced half-reactions for the redox reaction:
[tex]\[
2 NO_3^{-}(aq) + 2 H_2O(l) + 2 Cl^{-}(aq) \rightarrow 2 NO_2(g) + 4 OH^{-}(aq) + Cl_2(g)
\][/tex]
we should identify the oxidation states of the elements involved and separate the oxidation and reduction processes.
### Step 1: Identify oxidation states
- Nitrogen in [tex]\( NO_3^{-} \)[/tex] is +5.
- Nitrogen in [tex]\( NO_2 \)[/tex] is +3.
- Chlorine in [tex]\( Cl^{-} \)[/tex] is -1.
- Chlorine in [tex]\( Cl_2 \)[/tex] is 0.
### Step 2: Write and balance the half-reactions
#### Oxidation half-reaction:
Chlorine is oxidized from -1 to 0.
[tex]\[ 2 Cl^{-} \rightarrow Cl_2 + 2 e^{-} \][/tex]
#### Reduction half-reaction:
Nitrogen is reduced from +5 to +3.
[tex]\[ NO_3^{-} + 2 H_2O + 2 e^{-} \rightarrow NO_2 + 4 OH^{-} \][/tex]
### Step 3: Combine the half-reactions
Now, we ensure that the electrons lost in the oxidation half-reaction equal the electrons gained in the reduction half-reaction:
Oxidation half-reaction:
[tex]\[ 2 Cl^{-} \rightarrow Cl_2 + 2 e^{-} \][/tex]
Reduction half-reaction:
[tex]\[ NO_3^{-} + 2 H_2O + 2 e^{-} \rightarrow NO_2 + 4 OH^{-} \][/tex]
Since both half-reactions involve the transfer of 2 electrons, they can be directly combined:
### Balanced reaction:
[tex]\[ 2 NO_3^{-} + 2 H_2O + 2 Cl^{-} \rightarrow 2 NO_2 + 4 OH^{-} + Cl_2 \][/tex]
Therefore, the half-reactions for the given redox reaction are:
### Oxidation half-reaction:
[tex]\[ 2 Cl^{-} \rightarrow Cl_2 + 2 e^{-} \][/tex]
### Reduction half-reaction:
[tex]\[ NO_3^{-} + 2 H_2O + 2 e^{-} \rightarrow NO_2 + 4 OH^{-} \][/tex]
