Chemistry · Redox Equilibrium · NEET
Reducing power means how easily a metal gives away electrons (gets oxidised). A metal that gives electrons easily is a strong reducing agent. Two ways to check: (1) The activity series (K, Ca, Na, Mg, Al, Zn, Fe, ... Cu, Ag, Au). Metals at the top are the strongest reducers. (2) Standard reduction potential (E°). The MORE NEGATIVE the E°, the STRONGER the reducing agent. So K (very negative E°) is a much stronger reducer than Ag (positive E°). For NEET, if E° values are given, just line them up from most negative to most positive to get decreasing reducing power.
More negative E° = STRONGER reducing agent. This is the point students mix up most. A very negative E° means the metal really does not want to stay as an ion; it prefers to give away electrons and go back to the metal. Giving electrons = reducing the other substance = acting as a reducing agent. So a metal like K (E° = -2.93 V) is a strong reducing agent, while Ag (E° = +0.80 V) is a weak reducing agent but a good oxidising ion (Ag+ takes electrons easily).
They tell you the SAME thing in two forms. The activity series is a simple ranked list of metals by reactivity (how easily they lose electrons). Standard electrode potential (E°) is the exact number, in volts, behind that ranking. A metal with a more negative E° sits higher in the activity series. So the activity series is the 'ordered list' and E° is the 'measured value' that builds the list. In NEET, questions may give you either one and ask you to order reducing or oxidising power.
Oxidising power means how easily a species TAKES electrons (gets reduced). For halogens the order is F2 > Cl2 > Br2 > I2. Fluorine is the strongest oxidising agent because it grabs electrons most easily and has the most positive reduction potential. This is why a higher halogen can displace a lower one: Cl2 displaces Br- and I- from their salts, but I2 cannot displace Cl-. Memory: going DOWN the halogen group, oxidising power goes DOWN.
A more reactive metal is a stronger reducing agent, so it gives electrons more easily. When you put it with the ion of a less reactive metal, it pushes its electrons onto that ion and forces the ion back to metal. Example: Al + Cr2O3 -> Al2O3 + Cr (thermite reaction). Al is above Cr in the activity series, so Al displaces Cr. This is a displacement reaction and it directly shows the order of reducing strength.
Among common metals, the strongest reducing agent is one with the most negative E°, like Li or K. Among halogens (and common elements), the strongest oxidising agent is fluorine (F2), which has the most positive reduction potential. Simple rule: strong reducing agent = easily loses electrons (top of activity series); strong oxidising agent = easily gains electrons (F2).
The standard electrode potential (E°) values of Al3+/Al, Ag+/Ag, K+/K and Cr3+/Cr are -1.66 V, 0.80 V, -2.93 V and -0.74 V, respectively. The correct decreasing order of reducing power of the metals is
Which of the following reactions is a metal displacement reaction? Choose the right option.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
How easily it loses electrons. A metal high in the activity series, or with a more negative E°, loses electrons easily and is a strong reducing agent.
F2 > Cl2 > Br2 > I2. Fluorine is the strongest oxidising agent because it gains electrons most easily and has the most positive reduction potential.
There is a similar reactivity order for halogens based on oxidising power (F2 > Cl2 > Br2 > I2). The metal activity series ranks reducing power; the halogen order ranks oxidising power.
You are often given E° values and asked to order reducing or oxidising power. Line the values up: most negative E° = strongest reducing agent; most positive E° = strongest oxidising agent.
Yes. Zn is above Cu in the activity series, so Zn is the stronger reducing agent and displaces Cu: Zn + CuSO4 -> ZnSO4 + Cu.
