Chemistry · Redox Equilibrium · NEET
It is a single horizontal line for one element (like Br, Mn, or N). The species with the HIGHEST oxidation state is written on the far left, and the state drops as you move right, ending with the lowest state (often the free metal or the simple ion). The number written ABOVE each arrow is the standard reduction potential E° (in volts) for changing the left species into the right species. So each arrow is one reduction half-reaction with its voltage.
That E° is the standard reduction potential for the reduction going in the direction of the arrow (left species gaining electrons to become the right species). A more positive E° means that reduction happens more easily, so the left species is a stronger oxidising agent. This is why Latimer diagrams let you compare oxidising and reducing strength at a glance.
Pick the species you are asked about. Read the E° written just to its RIGHT (this is the couple where it gets REDUCED to a lower state). Then read the E° written just to its LEFT (this is the couple where it gets OXIDISED to a higher state). If E°(right) is GREATER than E°(left), the species disproportionates. In cell terms, E°cell = E°(right) − E°(left), and if that is positive the reaction is spontaneous.
Disproportionation means the SAME species is both oxidised and reduced at once. The right couple (higher E°) acts as the cathode (reduction) and the left couple (lower E°) acts as the anode (oxidation). E°cell = E°cathode − E°anode = E°(right) − E°(left). A positive E°cell means ΔG° is negative, so the reaction is spontaneous and the middle species splits into a higher and a lower state.
The species on the LEFT of the arrow with the most POSITIVE E° is the strongest oxidising agent (it grabs electrons easily). The species on the RIGHT of the arrow with the most NEGATIVE E° is the strongest reducing agent (it gives electrons easily). For NEET, this is the same logic as the electrode potential series, just drawn on one line.
Yes. E° values change with pH, so there are separate Latimer diagrams for acidic solution and basic solution. NEET questions will tell you the medium or give you the values directly, so just use the numbers printed on the diagram in the question. Do not mix acidic and basic values.
Consider the change in oxidation state of bromine corresponding to different E° values shown in the Latimer diagram: BrO4⁻ →(1.82 V) BrO3⁻ →(1.5 V) HBrO →(1.595 V) Br2 →(1.0652 V) Br⁻. The species undergoing disproportionation is:
Given: MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O with E° = +1.510 V (reduction); ½O2 + 2H⁺ + 2e⁻ → H2O with E° = +1.223 V. Will MnO4⁻ liberate O2 from water in the presence of acid?
The 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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The core idea (standard electrode potentials and predicting redox spontaneity) is squarely in the NEET redox and electrochemistry syllabus, and NEET 2018 asked a direct Latimer-diagram disproportionation question. So yes, learn to read one.
For any middle species, compute E°cell = E°(right arrow) − E°(left arrow). If it is POSITIVE, that species disproportionates. If it is negative or zero, it is stable and does not disproportionate.
The electrode potential series lists many different couples ranked by E°. A Latimer diagram shows only ONE element, laying out all its oxidation states in a row with the E° between each neighbouring pair. It is a compact map of a single element's redox behaviour.
Yes. You can combine arrows to find the E° between any two states, but you must weight by the number of electrons (E° is not simply averaged). For most NEET questions you only need the neighbouring arrows already printed on the diagram.
Yes. A more positive standard reduction potential means the species on the left of that arrow accepts electrons more easily, so it is a stronger oxidising agent. The species on the far right with the most negative E° is the strongest reducing agent.
