Question

Constants The following values may be useful when solving this tutorial. Constant Value E∘Cu 0.337 V E∘Zn -0.763 V R 8.314 J⋅mol−1⋅K−1 F 96,485 C/mol T 298 K Part A Part complete In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell are Cu2+(aq)+2e−→Cu(s) and Zn(s)→Zn2+(aq)+2e− The net reaction is Cu2+(aq)+Zn(s)→Cu(s)+Zn2+(aq) Use the given standard reduction potentials in your calculation as appropriate. Express your answer numerically to three significant figures. View Available Hint(s) -- SubmitPrevious Answers Keq K e q =

Answer 1

E°_cell = E°_cathode - E°_anode = 0.337V -(-0.763V)

= 1.100V

= 1.10V

E_cell = E°_cell - RT/nF × ln(K_eq) = 0

E°_cell = RT/nF × ln(K_eq)

1.10V = 8.314J/K.mol × 298K/(2 × 96485C/mol) × ln(K_eq)

ln(K_eq) = 1.10×96485×2/(298×8.314) = 85.6754

K_eq = e^85.6754 = 1.6157×10³⁷

= 1.62×10³⁷ (Answer)