Question

write cell half chemical reaction and global reaction:

PblPbSO4lSO42-llCu2+(aq)lCu
PtlTI+(aq),TI3+(aq)llCu2+(aq)lCu
CdlCdCl2llHNO3(aq), H2lPl
PtlH2(g)lHCl (aq)lCl2lPl

calculate delta G and Ks

Answer 1

T = 298 K

R = 8.314 J/mol-K

a)

PbSO4(s) + 2e^-- → Pb(s) + SO4^2-(aq)     E^o_red = -0.36 V

Cu²⁺(aq) + 2e- → Cu(s)       E^o_red = 0.34 V

E^o_cell = 0.34 – (-0.36) V = 0.7 V

ΔG = - nF E^o_cell = - 2 * 96485.3 * 0.7 = - 135079.5 J/mol

ΔG = - RT ln K

ln K = 135079.5 / (8.314 * 298) = 54.53

K = 4.77 x10²³

b)

Ti³⁺(aq) + 2e^-- → Ti⁺ (aq)     E^o_red = -1.03 V

Cu²⁺(aq) + 2e- → Cu(s)       E^o_red = 0.34 V

E^o_cell = 0.34 – (-1.03) V = 1.37 V

ΔG = - nF E^o_cell = - 2 * 96485.3 * 1.37 = - 264370 J/mol

ΔG = - RT ln K

ln K = -264370 / (8.314 * 298) = 106.7052

K = 2.2 x10⁴⁶

c)

Cd²⁺(aq) + 2e^- → Cd(s)     E^o_red = -0.4 V

2H⁺ + 2e^- → H2       E^o_red = 0.00 V

E^o_cell = 0.00 – (-0.4) V = 0.4 V

ΔG = - nF E^o_cell = - 2 * 96485.3 * 0.4 = -77188.3 J/mol

ΔG = - RT ln K

ln K = -77188.3 / (8.314 * 298) = 31.1548

K = 3.4 x10¹³

d)

2H⁺ + 2e^- → H2       E^o_red = 0.00 V

Cl2(g) + 2e- → 2Cl-(aq)       E^o_red = 1.36 V

E^o_cell = 1.36 – (0) V = 1.36 V

ΔG = - nF E^o_cell = - 2 * 96485.3 * 1.36 = -262440 J/mol

ΔG = - RT ln K

ln K = -262440 / (8.314 * 298) = 105.9263

K = 1.0 x10⁴⁶