A student builds a Cu concentration cell with two Cu electrodes immersed in 0.0010 M and 1.00 M Cu(NO3)2 with a salt bridge. At 25 C the measured cell potential is +0.088 V (left half-cell positive). E^o for the same metal in both half-cells is 0 V; the Nernst equation gives E_cell = -(0.0592/2) * log(Q).
What cell potential is predicted by the Nernst equation, and how does it compare to the measurement?
- Acheck_circle
E_calc = +0.089 V; agrees with the +0.088 V measurement
- B
E_calc = +0.18 V; cell is non-ideal
- C
E_calc = 0 V; the cell does not operate
- D
E_calc = -0.089 V; cell is reversed
Explanation
Q = [Cu^2+]_dilute / [Cu^2+]_conc = 0.001/1.00 = 1e-3. E = -(0.0592/2) log(1e-3) = -(0.0296)(-3) = +0.0888 V. Matches measurement.