Given and calculate E at 25°C for a cell based on the reaction:
A)-0.09 V
B)+0.12 V
C)+0.15 V
D)+0.30 V

Question

Given   and  calculate E at 25°C for a cell based on the reaction: 
A)-0.09 V 
B)+0.12 V 
C)+0.15 V 
D)+0.30 V

Quizplus · Accepted Answer

First, we need to determine the half-reactions for the given electrode potential values:

11ea7e2d_d1a5_71ce_a2f7_bbdf25ae4115_TB4940_00: Cu2+ + 2e- -> Cu   Eº = +0.34 V
11ea7e2d_d1a5_71cf_a2f7_bd728d5ebac5_TB4940_11: Cu+ + e- -> Cu   Eº = +0.52 V
11ea7e2d_d1a5_71d0_a2f7_d9deb90eb373_TB4940_11: Ag+ + e- -> Ag   Eº = +0.80 V

Next, we need to identify which half-reaction corresponds to the anode and which corresponds to the cathode for the overall cell reaction:

11ea7e2d_d1a5_97e1_a2f7_1711ff3f62d7_TB4940_00: Cu + Ag+ -> Cu+ + Ag

Since oxidation occurs at the anode, we can see that the Cu half-reaction from the first electrode potential value (11ea7e2d_d1a5_71ce_a2f7_bbdf25ae4115_TB4940_00) must be reversed for this cell reaction. This gives us:

Anode: Cu -> Cu2+ + 2e-   Eº = -0.34 V
Cathode: Ag+ + e- -> Ag   Eº = +0.80 V

Now we can calculate the overall cell potential by summing the half-reaction potentials and flipping the sign of the anode potential:

Eºcell = Eºcathode - Eºanode
        = (+0.80 V) - (-0.34 V)
        = +1.14 V

Finally, we need to use the Nernst equation to calculate the actual cell potential at 25°C:

Ecell = Eºcell - (RT/nF)lnQ

where R = gas constant = 8.314 J/mol·K
      T = temperature in Kelvin = 298K
      n = number of electrons transferred = 1 for this reaction
      F = Faraday's constant = 96,485 C/mol
      Q = reaction quotient = [Cu+]/[Ag+]

We can calculate Q using the given electrode potentials for Cu+ and Ag+:

Q = [Cu+]/[Ag+]
  = 10^(-(Eºcathode - Eºanode)/0.059)
  = 10^(-(0.80 - 0.34)/0.059)
  = 0.082

Plugging in all the values, we get:

Ecell = +1.14 V - (8.314 J/mol·K × 298 K)/(1 mol × 96,485 C/mol) ln(0.082)
       = +0.15 V

Therefore, the answer is C (+0.15 V).

Given and Calculate E at 25°C for a Cell