Electrochem problem....

Question

The cell

C(gr) I Cl2(g) I Cl- (aq) II Au^3 (aq) I Au(s)

is under conditions such that the measured potential is 0 V. The standard cell emf is 1.4 V at the cathode and 1.36 V at the anode. Calculate the reaction quotient Q for the cell reaction under these conditions

Answer 1

2 Cl-  --->  Cl2 & 2 e-  (the oxidation ...Eo = - 1.36 volts)

Au+3 &  3 e-  -->  Au   ( the reduction ... Eo = 1.4 volts)

those voltages give an Eo cell = + 0.04 volts

(double check the Au voltage & see if it isn't actually 1.49 to 1.50)

nernst:

E = Eo - (0.0592 / n) (log Q)

0 = 0.04 - (0.0592 / 6) (log Q)

0.04 = (0.0592 / 6) (log Q)

0.04 / 0.00987 = (log Q)

log Q = 4.054

Q = 11,325

unless there are more digits with the voltage of the gold cell, there are almost no sig figs in this problem

check that voltage out, redo if you have to, I'm leaving rounding off to you