Use the standard half-cell potentials listed below to calculate the standard cel

Question

Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 degree C. (The equation is balanced.) Sn(s) + 2 Ag^-(aq) rightarrow Sn^2+(aq) + 2 Ag(s) Sn^2+ (aq) + 2e^- rightarrow Sn(s) E degree = -0.14 V Ag^+(aq) + e^- rightarrow Ag(s) E degree = +0.80 V A) +0.94 V B) +1.74 V C) -1.08 V D) +1.08 V E) -1.74 V Use the tabulated half-cell potentials to calculate delta G for the following balanced redox reaction. Pb^2+(aq) + Cu(s) rightarrow Pb(s) + Cu^2 + (aq) A) -0.47 kJ B) +91 kJ C) -41 kJ D) -21 kJ E) +46 kJ Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25 degree C. 3I_2(s) + 2 Fe(s) righarrow 2 Fe^3+(aq) + 61^-(aq) A) 2.4 times 10^58 B) 8.9 times 10^-18 C) 1.1 times 10^17 D) 3.5 times 10^-59 E) 1.7 times 10^29 Determine the cell notation for the redox reaction given below. Sn(

Explanation / Answer

The half having higher reduction potential undergoes reduction at cathode and the other undergoes oxidation at anode.

(17)

E0cell = E0Ag+/Ag - E0Sn2+/Sn = + 0.80 - ( - 0.14 ) = + ).94 V

(A)

(18)

E0cell = E0Pb2+/Pb - E0Cu2+/Cu = - 0.126 - 0.342 = - 0.468 V

deltaG0 = - n F E0cell = - 2 * 96500 * ( - 0.468 ) = + 90324 J = + 91 kJ

(B)

(19)

E0cell = E0I2/I- - E0Fe3+/Fe = 0.536 - ( - 0.0400) = + 0.576 V

DeltaG0 = - 6 * 96500 * 0.576 = - 333504 J

but, DeltaG0 = - R T lnK

- 333504 = - 8.314 * 298 * lnK

lnK = 134.61

K = 2.88 * 1058

(A)

(20) (C)

Oxidation half folloed by reduction half in which both are seperated by salt bridge. (two vertical lines)

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

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