Separate this redox reaction into its component half-reactions.
Cl2 + 2Na --> 2NaCl
is it
e- + Cl2 = 2Cl^-
Redox reaction
What is being oxidised and reduced
To separate this redox reaction into its component half-reactions, you need to identify the species that is being oxidized (reducing agent) and the species being reduced (oxidizing agent). In this case, Cl2 is being reduced to form NaCl.
Let's start with the oxidation half-reaction:
Cl2 --> 2Cl-
To balance the number of chlorine atoms on both sides, we need to add 2 electrons (e-) to the left-hand side:
Cl2 + 2e- --> 2Cl-
Now, let's move to the reduction half-reaction:
2Na+ + 2e- --> 2Na
To balance the charges, we need to add 2 sodium ions (Na+) on the left-hand side:
2Na+ + 2e- --> 2Na
Now we have identified the oxidation half-reaction as Cl2 --> 2Cl- and the reduction half-reaction as 2Na+ + 2e- --> 2Na.
Na ==> Na+
Cl2 ==> Cl^-
You balance
close
2e + Cl2 ==> 2Cl^-
In a redox equation it must balance three ways.
1. by atoms
2. by charge
3. by electron change.
Na ==> Na^+ + e
Cl2 + 2e ==> 2Cl^-
Multiply #1 by 2 and #2 by 1 to make the electron change the same.
2Na ==> 2Na+ + 2e
2e + Cl2 ==> 2Cl^- and add
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2Na + 2e + Cl2 ==> 2Na^+ + 2e + 2Cl^-
The 2e cancel (that's why you make them equal; they are on opposite sides and will cancel). You are left with
2Na + Cl2 ==> 2NaCl