Answer:
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.
; where the "M" stands for molarity, the "mol" stands for moles of solute and the "liters" means the volume in liters of solution.
= 0.5M