Question

The concentrations of NaCl and KI are at 0.150 M and 0.100 M, respectively, in the following electrochemical cell:

A) Using the following half-reactions, calculate cell voltage:

CuI(s) e- ----> Cu(s)I- Eo=-.185

AgCl(s) +e- ----> AgCl- Eo=.222

E=?

hint:Part A is solved by calculating the half-cell potential for the right and left half-cell. Use the Nernst equation and the concentration of the counter ion (Cl– or I–) to find each half-cell potential. There is only one electron involved in these reactions. Part B is solved differently, as the half-cell reaction does not contain only solid metal (Ag or Cu) – it contains some in solution. Therefore, you must use the solubility product constant (Ksp) to determine the concentration of the free Ag and Cu in solution. Then, use the Nernst equation with the correct standard reduction potential for the half reactions and the free metal ion concentration you just determined to find the half-cell voltage. Subtract the left half-cell voltage from the right half-cell voltage to find the total cell voltage.

struggling here, thanks for the help

Answer 1

For part A, we first calculate half-cell potentials with nernst equation:

E = E^o - 0.0592/n log Q

For Copper half cell:

E = -0.185 - 0.0592/1 log (1 / 0.1)

E = -0.2442 V

For Silver Half cell:

E = 0.222 - 0.0592/1 log (1 / 0.15)

E = 0.1732 V

Now, we obtain global voltage:

E_cell = E_cathode + E_anode

We then determine which is which, as we can see, silver half-cell has the higher voltage, so we can say that is the reduction (cathode) and copper cell is oxidation (anode), so we must change the sign of oxidation voltage, assuming standard voltages given are reduction potentials:

E_cell = E_cathode + E_anode

E_cell = 0.1732 + 0.2442 = 0.4174 V