Given the half‑reactions and their respective standard reduction potentials 1. Cr 3 + + e − ⟶ Cr 2 + E ∘ 1 = − 0.407 V 2. Cr 2 + + 2 e − ⟶ Cr ( s ) E ∘ 2 = − 0.913 V calculate the standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s).

The standard reduction potential for the reduction half‑reaction of Cr(III) to Cr(s) is -0.744 V

Explanation:

Here we have

1. Cr³⁺  + e − ⟶ Cr²⁺ E⁰₁ = − 0.407 V

2. Cr²⁺ + 2 e − ⟶ Cr ( s ) E⁰₂  = − 0.913 V

To solve the question, we convert, the E⁰ values to ΔG as follows

ΔG₁ = n·F·E⁰₁ and ΔG₂ = n·F·E⁰₂

Where:

F = Faraday's constant in calories

n = Number of e⁻

ΔG₁ = Gibbs free energy for the first reaction

ΔG₂ = Gibbs free energy for the second half reaction

E⁰₁  = Reduction potential for the first half reaction

E⁰₂ = Reduction potential for the second half reaction

∴ ΔG₁ = 1 × F × − 0.407 V

ΔG₂ = 2 × F  × − 0.913 V

ΔG₁  + ΔG₂  = F × -2.233 V which gives

ΔG = n × F × ΔE⁰ = F × -2.233 V  

Where n = total number of electrons ⇒ 1·e⁻ + 2·e⁻ = 3·e⁻ = 3 electrons

We have, 3 × F × ΔE⁰ = F × -2.233 V

Which gives ΔE⁰ = -2.233 V /3 = -0.744 V.

- aluminum and copper chloride.

- the chloride ion to separate the aluminum from the oxygen so that the aluminum can react with the copper ions (and the water molecules).

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