Question

Use the References to access important values if needed for this question. When the Pb2+ concentration is 1.41 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.170V. What is the Mn2+ concentration? Pb(s) + Mn2+(aq) Pb2+ (aq) + Mn(s) — Answer: M Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

Pb2+ (aq) + n(s) Pb (w + Mn2+ cor) ...) Reduction half reaction: Pb2+ + 2ē -> Pb(3) Ereq = -0.13V oxidation half reaction: mm () - MM2+62) + 2ē E ºox = Ered =- (119) 1.19 intial Strandard - Eired + Bºox = 6013) + (119) = 1:06 V Reaction quotient of reaction (;) m2+] P t Pb2+] from Nernst equation we get E cell - E cell - RT en Q.... (in) where, R=Gan constant = 8.314 I morki T= temperature in kelvin = 2985 = Number of electrons involved F = Faraday constant = 96500 t mol mat P- Company, [P624) = ;-, M [ Pb2th E cell = 1.065, 1170v (given) 2+ - 1.4iM

Pugging all these value into equation (in) 8:314 X 298 1170 - 106 - 2 x 96500 =) 1170 .- 1.06 - 00128 ln & =) 0.0128bn Q = "06 - 1170 = -0') en & - – O'll soon8 = -8.594 → Qué 8.894 185 x10-4 EPb2th = 1.85x18-4 1. # I unat] = 1.41 X1. 85X10-9 [romat) = 2:6x18" en (Answer) 3Pb2+ + 2 er 6) —> 36PB)+2 cm++ (ar.) ...fü) Reduction half reaction. apb2+ a + b 30 b (s) ; Elle oxidation has f reaction: 2 cris) - 2 Cr3+ (aq) + bé; Elox = -03V en Ered - (074) = 0.740 . Ell- Eox & Ered = 0.74 + 0.13) = 0.61 v ... Rention quotient of the reaction (ii) a= [cat/[Pbaty 3

here, & cell = 0.5177 (given) using earration (uj) = 67 0.517 = 00 - 8.3/4 X 298 6 x 96500 le Q = 0.517 = 0.61 – 0.00028 ln & = 0.00428 ln Q = 0.61 -0.517 - 0 093 en Q = 0.093 = 21.729 b.d0428 = 21.729. EPb2+3 (6.90x104)3 = 21.729 =) [cr2+)?? 21-729(6-9 0X1024) => [Cr3+-7138 X10-9 .: [Cr3+) = V 7.138 X10-9 = 8.45 x10-5m . [crs] = 8.45x10 - es (Answer)