Question

If a platinum indicator electrode is used to measure [Cu²⁺] (standard reduction potential of Cu²⁺ is 0.337 V), and [Cu²⁺] = 0.0053 M, what is the cell potential (V) if the reference electrode is SCE and the measurement is taken at 298 K?

I have gotten .270 and .073, Im not sure what I am doing wrong

Answer 1

note : you are correct.0.270 V is correct. i do not find any mistake in your answer. same i show you again

anode : H2 (1 atm) -------------------> 2H+ (1 .00 M ) + 2e-    , Eo = 0.00V

cathode : Cu2+ + 2e- ------------------> Cu                              , Eo = 0.337 V

overall reaction : H2 + Cu2+ ------------> Cu + 2H+    , Eocell = Eo cathode - Eo anode

                                                                                                   = 0.337 - 0.00

                                                                                                   = 0.337 V

Q = [H+]^2 / PH2 x [Cu2+]

Q = (1)^2 / 1 x 0.0053

Q = 188.7

Ecell = Eocell - 0.05916 / n * log Q

           = 0.337 - 0.05916/2 * log 188.7

            = 0.270 V