you added 0.05mL of 0.10M AgNO3 to 4.0mL of 2.0M NaCl and formed a saturated solution of AgCl (s): AgCl(s) -> Ag+(aq)+Cl-(aq). Using this saturated solution as the test solution you set up the following cell: Ag|Ag+(test solution after reaction||Ag+(1.0M)|Ag and measured Ecell=0.61V. Using the measured value of Ecell calculate the equilibrium concentration of Ag+ (i.e. of the test solution after reaction).
First, let's write out the balanced half-reactions for the cell:
Anode: Ag(s) -> Ag+(aq) + e-
Cathode: Ag+(1.0M) + e- -> Ag(s)
From the given Ecell value of 0.61V, we can use the Nernst equation to calculate the equilibrium concentrations of Ag+:
Ecell = Ecathode - Eanode
0.61V = EAg+1.0M - EAg
0.61V = 0 - 0.80V
0.61V = -0.80V
Using the Nernst equation:
Ecell = E°cell - (RT/nF)lnQ
0.61V = 0.80V - (0.0257V)log(Q)
Rearranging the equation:
ln(Q) = (0.80V - 0.61V) / (0.0257V)
ln(Q) = 7.41
Q = e^7.41
Q = 1636.82
Now we can calculate the concentration of Ag+ in the test solution after the reaction:
Q = [Ag+]/[Ag+(1.0M)]
1636.82 = [Ag+]/(1.0M)
[Ag+] = 1.0M * 1636.82
[Ag+] = 1636.82M
Therefore, the equilibrium concentration of Ag+ in the test solution after the reaction is 1636.82M.