The Activity Series
- Metals can be ranked in order of reactivity -this is called the reactivity series or just activity series
- Metals higher in reactivity can displace less reactive metals from their compounds in solutions or from their oxides
- An example of a metal displacement reaction occurs between magnesium and aqueous copper (II) sulfate solution
Mg (s) + CuSO4 (aq) → MgSO4 (aq) + Cu (s)
- What are the redox processes going on here? If we split the equation into half equations it’s easy to see which species is oxidized and which is reduced:
Mg → Mg2++ 2e- Loss of electrons = oxidation
Cu2+ + 2e- → Cu Gain of electrons = reduction
- The more reactive metal acts as a reducing agent
- This allows metals to be ranked from most reactive (strongest reducing agents) to least reactive:
The more reactive a metal is the better it is at pushing electrons onto less reactive metal ions. Magnesium is better at pushing electrons onto copper(II) ions than copper is at pushing electrons onto magnesium ions
Worked Example
What is the order of decreasing reactivity of the metals (most reactive first)?
Zn(s) + Sn2+(aq) → Zn2+(aq) + Sn(s)
Cu(s) + Zn2+(aq) → No ReactionSn(s) + Cu2+(aq) → Sn2+(aq) + Cu(s)
Ag(s) + Cu2+(aq) → No ReactionA. Zn > Cu > Sn > Ag
B. Sn > Zn > Ag > Cu
C. Ag > Cu > Zn > Sn
D. Zn > Sn > Cu > Ag
Answer:
The correct option is D.
- The first reaction tells you that zinc is more reactive than tin (Zn > Sn):
Zn(s) + Sn2+(aq) → Zn2+(aq) + Sn(s)
- The second reaction tells you that zinc is more reactive than copper (Zn > Cu):
Cu(s) + Zn2+(aq) → No Reaction
- The third reaction tells you that tin is more reactive than copper (Sn > Cu):
Sn(s) + Cu2+(aq) → Sn2+(aq) + Cu(s)
- The fourth reaction tells you that copper is more reactive than silver (Cu > Ag):
Ag(s) + Cu2+(aq) → No Reaction
Deducing the Feasibility of a Redox Reaction
- Given the activity series of metals it is possible to predict whether a displacement reaction will take place or not ( if it is feasible)
- For example will calcium displace lead from lead(IV)oxide?
PbO2 + 2Ca → Pb + 2CaO
- Deduce the redox half equations
Ca → Ca2+ + 2e-
Pb4+ + 4e- → Pb
- Ca is above Pb in the activity series, so this reaction is feasible
- The more reactive metal undergoes oxidation (Ca in this case)
- The less reactive metal ion undergoes reduction
Exam Tip
You don't need to learn the Activity Series for Metals as it is given to you in Section 25 of the IB Chemistry Data Booklet