In an electrochemical reaction of lead, at standard temperature, if E⁰( Pb²⁺ / Pb)=m Volt and E⁰( Pb⁴⁺ / Pb)=n Volt, then the value of E⁰( Pb²⁺ / Pb⁴⁺) is given by m-x n. The value of x is ___________. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Potentials do not add, but free energies do: go through Δ G° = -nFE°.

1. Write both couples. Pb²⁺ + 2e⁻ → Pb has E° = m, so Δ G₁° = -2Fm. Pb⁴⁺ + 4e⁻ → Pb has E° = n, so Δ G₂° = -4Fn.
2. Build the target. Subtract the first reaction from the second. This gives Pb⁴⁺ + 2e⁻ → Pb²⁺, a 2-electron change, and its Δ G° is Δ G₂° - Δ G₁°.

   -2F E°_(Pb⁴⁺/Pb²⁺) = -4Fn - (-2Fm) ⇒ E°_(Pb⁴⁺/Pb²⁺) = 2n - m

3. Reverse the direction. The question asks for Pb²⁺/Pb⁴⁺, the change Pb²⁺ → Pb⁴⁺ + 2e⁻. Reversing a reaction flips the sign of its potential, so E° = -(2n - m) = m - 2n.
4. So. Comparing m - 2n with m - xn gives x = 2.

The common slip: 1. Subtracting the two potentials directly, without weighting each by its electron count, gives m - n, so x = 1.

The idea to keep. To get a new electrode potential from two others, convert each to Δ G° = -nFE°, add or subtract the Δ G° values, then divide by the new electron count.

Answer from the NTA final answer key. Solution by PrepWiser.