For the following electrochemical cell Cu(s)|Cu2 (aq, 0.0155 M)||Ag (aq, 1.50 M)
ID: 998651 • Letter: F
Question
For the following electrochemical cell Cu(s)|Cu2 (aq, 0.0155 M)||Ag (aq, 1.50 M)|Ag(s) write the net cell equation
Question 14 of 15 tr Mapco eneral Chemis Donald McQuarrie Peter A. Rock Ethan Gallogly University Science Books presented by Sapling Learning For the following electrochemical cell Cu(s)Cu (aq, 0.0155 M)lIAg (aq, 1.50 M)lAg(s) write the net cell equation. Phases are optional. Do not include the concentrations. Calculate the following values at 25.0 °C using standard potentials as needed. Number Number k.J/ mol cell Number Number kJ/ mol cellExplanation / Answer
Cell reaction is:
Cu(s) + 2 Ag+(aq) => Cu2+(aq) + 2 Ag(s)
Ag+ + 1e- >> Ag E° = 0.800 V
Cu2+ + 2e- >> Cu E° = 0.337 V
for the half-reaction
Cu >> Cu2+ + 2e- E° = - 0.337 V
total reaction that will occur
2 Ag + Cu >> 2Ag + + Cu2+ E cell = 0.800- 0.337 = 0.463 V
Eo(cell) = 0.463 V
in this reaction , Moles of electrons transferred = n = 2
Faraday constant F = 96485 C/mol
Temperature T = 25 deg C = 298.15 K
Molar gas constant R = 8.314 J/mol.K
Nernst equation:
E(cell) = Eo - RT/nF ln([Cu2+]/[Ag+]^2)
= 0.463 - 8.314 x 298.15/(2 x 96485) x ln(0.0155/1.50^2)
= 0.463 - 8.314 x 298.15/(2 x 96485) x (-4.98)
= 0.526 V
now calculate the Delta Go(rxn) as follows:
Delta Go(rxn) = -nFEo(cell)
= -2 x 96485 x 0.463V
= -8.93 x 10^4 J/mol
= -89.3 kJ/mol
Delta G(rxn) = -nFE(cell)
= -2 x 96485 x 0.526 V
= -1.02 x 10^5 J/mol = -102 kJ/mol