Examine the unbalanced electrolytic reaction and its two half-reactions.
Cd2+(aq) + Br2(g) → Cd(s) + Br–(aq)
E0 = –1.49 V
Half-reactions:
Cd2+(aq) + 2e– → Cd(s)
Br2 (g) → Br–(aq) + e–
What is the balanced form of this reaction?
Cd2+(aq) + 2Br2(g) → Cd(s) + 4Br–(aq)
E0 = –1.49 V
3Cd2+(aq) + 2Br2(g) → 3Cd(s) + 4Br–(aq)
E0 = –1.49 V
2Cd2+(aq) + Br2(g) → 2Cd(s) + 4Br–(aq)
E0 = –1.49 V
Cd2+(aq) + Br2(g) → Cd(s) + 2Br–(aq)
E0 = –1.49 V
The balanced form of the reaction is:
2Cd2+(aq) + Br2(g) → 2Cd(s) + 4Br–(aq)
To find the balanced form of the given reaction, we need to balance the number of atoms of each element on both sides of the equation.
First, let's balance the Cd atoms:
On the left side, there is 1 Cd atom. On the right side, there is also 1 Cd atom.
So the Cd atoms are already balanced.
Next, let's balance the Br atoms:
On the left side, there are 2 Br atoms in Br2. On the right side, there are 4 Br atoms in 4Br-.
To balance the Br atoms, we need to multiply the left side by 2:
Cd2+(aq) + 2Br2(g) → Cd(s) + 4Br-(aq)
So, the balanced form of the reaction is:
Cd2+(aq) + 2Br2(g) → Cd(s) + 4Br-(aq)
The value of E0, which is -1.49 V, remains the same for the balanced reaction.