Bonjour,
1) Ag⁺ + e⁻ → Ag
Cu²⁺ + 2e⁻ → Cu
2) 2Ag⁺ + Cu → 2Ag + Cu²⁺
3)
2Ag⁺ + Cu → 2Ag + Cu²⁺
état avanct.
initial 0 ni(Ag⁺) ni(Cu) 0 0
en cours x ni(Ag⁺) - 2x ni(Cu) - x 2x x
final xf ni(Ag⁺) - 2xf ni(Cu) - xf 2xf xf
4) ni(Ag⁺) = c x V = 0,200 x 50,0.10⁻³ = 1,00.10⁻² mol
ni(Cu) = m/M(Cu) = 1,00/63,5 ≈ 1,57.10⁻² mol
5)
ni(Ag⁺) - 2xf = 0 OU ni(Cu) - xf
⇒ xf = ni(Ag⁺)/2 = 5,00.10⁻³ mol OU xf = ni(Cu) = 1,57.10⁻² mol
⇒ xmax = 5,00.10⁻³ mol
Ag⁺ est limitant
6) Etat final
nf(Ag⁺) = ni(Ag⁺) - 2xmax = 0 mol
nf(Cu) = ni(Cu) - xmax = 1,57.10⁻² - 5,00.10⁻³ = 1,07.10⁻² mol
nf(Ag) = 2xmax = 1,00.10⁻² mol
nf(Cu²⁺) = xmax = 5,00.10⁻³ mol
7) il faut : ni'(Ag⁺) = 2 x ni(Cu) = 2 x 1,57.10⁻² = 3,15.10⁻² mol
Soit : V' = ni'(Ag⁺)/c = 3,15.10⁻²/0,200 ≈ 157 mL
8) m(Ag) = nf(Ag) x M(Ag)
avec nf(Ag) = 2xmax = 2 x 1,57.10⁻² mol = 3,15.10⁻² mol
soit : m(Ag) = 3,15.10⁻² x 108 = 3,40 g
