Question

5. (a) Use the standard reduction potentials at 25° C in Table 18.1 in Tro, Fridgen and Shaw, and calculate the standard emf Eº of an electrochemical cell described by the following reaction: Pb + Sn2+ --> Pb2+ + Sn (b) What is the value of E if the concentration of Sn is 1.26 x 10-2 and the concentration of Pb2+ is 7.11 x 10-5? (c) If E = 0.16 V and Sn2+ = 0.10 M, what is the concentration of Pb2+?

Answer 1

(5) Giren reaction (a) pb + Sn 27 Pb2+ + Sn - from the table 18.1, pbeltzee -> Pb Epleton = -0.13 y → Embl0b2t- +0.13 Y Son 2+ + 2e → Sn Eu 2+/80 = -0.14V In the given 8m 0, pb undergoes onide & Sozimdugoes reductim .. E cele = Standard emt for - Ep/p12+ + Ep24/80 = 0.13 – 0.14V Eiell. =-0.011. (b) Giren [sn 2+] = 1126810² [p52+] = 711x105 from Neenest tam, Ecen= tcell nF . [Pb2 [sn] na no- gelectrons RT 1 ] tromsferred... [[sn] = lumit [pb] = lunit] 0.059 = C ph] ( sn 2t [Pb2+] [sn 27] 7.41x155 Ecell- =-0.01 - aux log - 12 26 1152 = -0.01 – 0.0591. xlog (5.643816-3) 2-0.01-0.02955% (+2.25)

Ecele = -0.01 + 0.06644 Ecell = 0.05644 V - (C) Given when E= 0.16 V , & [Sn 2+] = 0.1om then [Pb2+] = ? Fell - 0.0591 log [Pb2+] ► 0.16V= -.01 -0.05af1og [Pb213 –log Cmary] = 0.171 = -0.02955«[109 [Pb2+)- 109 (0.1)] ► loglio it. 0.02955V → 109[p524) = -5.7529-1 3-6.7259 = (pb24) = 106.7259 => [Pb2+] = 1.88x10 - 7 mm i concentratim 9 (Pb2+] = 1,88x10mm
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