the reaction Cr | Cr3+ || ZN2+ | Zn is at equilibrium at 25*Cusing the standard

Question

the reaction Cr | Cr3+ || ZN2+ | Zn is at equilibrium at 25*Cusing the standard voltage, what is the concentration of Zn2+ when[Cr3+] equals .19M i got up to this so far: E= -.018 -(8.314*298) / 6e(96485) LN .19^2/X^3 did i do it right help! the reaction Cr | Cr3+ || ZN2+ | Zn is at equilibrium at 25*Cusing the standard voltage, what is the concentration of Zn2+ when[Cr3+] equals .19M i got up to this so far: E= -.018 -(8.314*298) / 6e(96485) LN .19^2/X^3 did i do it right help! i got up to this so far: E= -.018 -(8.314*298) / 6e(96485) LN .19^2/X^3 did i do it right help!

Explanation / Answer

(Cr3+ + 3e   ->Cr(s))*2              E_half =   0.74 V (Zn(s)           -> Zn2+ + 2e)*3   E_half = +0.7618V ---------------------------------------- 2Cr3+ + 3Zn(s)      2Cr(s) +3Zn2+ E0 = 0.7618 - 0.74 = 0.0218 V delG0 = -nFE0,        n = 6electrons (in order to eliminate electron from the overallequation) ,   F = 96485 C/mol delG0 = -nFE0 = -RTlnK K = exp(nFE0/RT) = exp(6*96485 *0.0218/(8.314 *298)) =163.006917 K = [Zn2+]3 /[Cr3+]2 [Zn2+] = (K *[Cr3+]2)1/3 =(163.00691*0.192)1/3 = 1.81 M

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.