Question

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 0.0530 M and 1.60 M , respectively.

1. What is the initial cell potential?   

2. What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M ?

3. What is the concentration of Pb2+ when the cell potential falls to 0.370 V ?

Answer 1

0 The cell is represented as i pb, pb2t (0.053 m) || cat, cu (1.6 m) Emf of the cell E = 8°(0.34-6-0,127)) - 0.059) loglouett er n= l as zē are involved in the redne procen. ES 0.467 V -- 0.053 olosul log T.6 0.85107 V Autolific toward inter med when the conc. of culf has fallen to 0.2m ze 1.6 20.2 = 1.4m has been censunted ph must have been produced so rcult] = 0.2mingo a Front [062+] = (1.4 +0.0533) 31.453 M E = 0.467 - 0.0519 long halos Ei = 8.4410 © E = 0.370 v, [Pb2+] + [62743 = 16 701053 E = 0.3700 = 0.4670- og log (p. 47 -9 longe 7037 -0.469) *2 3.282 = 1.653m - 6 0105 91

Срь 1) . 141. 2 ( +) (рь+ ) . I {1 с. +7 (4+) putting this in eq 20 [Pb2+] + [au24]= 1.653 m г! Ср, 2 ] + Crb F) . 33 1916. }} (1912. 2+) CPь?t) = | 6 5 5 x 195 % - 2 Ср+) - 1 653 x 19 | 22 • 1912. Се 22 C++ ) = 1.65 21 и