Finding E°cell using Latimer Diagrams: The Ultimate Shortcut
If you are given the Standard Reduction Potential (E°) of Cu2+ → Cu+ and Cu+ → Cu, and you need to find the E° of Cu2+ → Cu, what do you do?
Why? Standard Reduction Potential (E°) is an intensive property. It does not depend on the amount of substance and cannot be added algebraically. You must use Gibbs Free Energy (ΔG°), which is an extensive property.
To save time in JEE and NEET exams, we use a structural approach called a Latimer Diagram combined with a direct shortcut formula derived from ΔG°.
Video Tutorial: The Latimer Diagram Trick
Watch Abhishek Sengar sir explain how to bypass the lengthy ΔG° derivation and use the shortcut formula to solve this in under 30 seconds!
The Latimer Diagram Shortcut
First, write the species in order of decreasing oxidation state. For our Copper example: Cu2+ → Cu+ → Cu.
Fig: Latimer Diagram for Copper. n represents the number of electrons transferred.
The shortcut formula relies on the fact that ΔG°3 = ΔG°1 + ΔG°2. Since ΔG° = -nFE°, the Faraday constant (F) and the negative signs cancel out, leaving us with the master equation:
Where n3 = n1 + n2
- For Cu2+ → Cu+: Electrons exchanged (n1) = 1. Potential = x.
- For Cu+ → Cu: Electrons exchanged (n2) = 1. Potential = y.
- For overall Cu2+ → Cu: Electrons exchanged (n3) = 2.
- Apply Formula: E°3 = [1(x) + 1(y)] / 2 = (x + y) / 2 V.
Practice Questions for JEE & NEET
Verify your understanding with these conceptual and numerical problems.
Question 1: Given E° for Fe3+ → Fe2+ is 0.77 V and E° for Fe2+ → Fe is -0.44 V. Calculate the standard reduction potential (E°) for Fe3+ → Fe.
Answer: -0.036 V
Step-by-step Solution:
- Step 1: Fe3+ → Fe2+. n1 = 1, E°1 = 0.77 V
- Step 2: Fe2+ → Fe. n2 = 2, E°2 = -0.44 V
- Overall: Fe3+ → Fe. n3 = 3 (since 1 + 2 = 3)
- Apply Formula: E°3 = ( n1E°1 + n2E°2 ) / n3
- E°3 = ( 1 × 0.77 + 2 × -0.44 ) / 3
- E°3 = ( 0.77 - 0.88 ) / 3 = -0.11 / 3
- E°3 = -0.0367 V
Question 2: Conceptual: We know we cannot add E° values directly because it is an intensive property. But when we calculate the E°cell of a full Galvanic cell using the formula E°cell = E°cathode - E°anode, we are essentially adding/subtracting E° values. Why is this allowed?
Answer: Because the number of electrons transferred (n) is the same for both half-reactions in a balanced Galvanic cell.
When you combine an oxidation half-reaction with a reduction half-reaction to form a complete cell, you mathematically balance the electrons so that n cancels out entirely.
In Latimer diagrams (like the one above), we are combining two consecutive reduction reactions to find a third reduction reaction. The number of electrons (n) accumulates instead of canceling out, so you MUST use the ΔG° pathway to account for the changing n values.
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