In the balanced equation 2 Al(s) + 3 Ni2+(aq) → 2 Al3+(aq) + 3 Ni(s), what is the coefficient of Al(s)?

• A. 1
• B. 2
• C. 3
• D. 4

Answer: (B)

Balancing a redox reaction by equalizing the electrons transferred between oxidation and reduction is the idea here. Aluminum metal is oxidized: Al(s) → Al3+(aq) + 3 e−. Nickel(II) ions are reduced: Ni2+(aq) + 2 e− → Ni(s). To make the electrons match, multiply the oxidation half-reaction by 2, giving 2 Al(s) → 2 Al3+(aq) + 6 e−, and multiply the reduction half-reaction by 3, giving 3 Ni2+(aq) + 6 e− → 3 Ni(s). Adding these yields the balanced overall equation: 2 Al(s) + 3 Ni2+(aq) → 2 Al3+(aq) + 3 Ni(s). So the coefficient in front of Al(s) is 2. If you tried 1, the electron transfer wouldn’t balance with an integer number of Ni2+ reductions, hence you need 2 aluminum atoms to pair with 3 nickel(II) ions.