Will you please help me figure out this chemistry problem?

An electrode is prepared from liquid mercury in contact with a saturated solution of mercury(I) chloride, Hg2Cl2, containing 1.00 M Cl-. The emf of the voltaic cell constructed by connecting this electrode as the cathode to the standard hydrogen half-cell as the anode is 0.268 V. What is the solubility product of mercury(I) chloride?

Will you please help me figure out this chemistry problem?
I find the following data from Wikipedia:


Hg2(2+)(aq) + 2e− → 2Hg(l), Eº = +0.80V


From:


E = 0.268 = Eº - RT/nF Ln Q


we have:


1/Q = [Hg2(2+)] = exp(-(Eº-E)nF/RT)


= exp(-0.532*2*96485/(8.314*298)) = 1.01x10^-18


Since [Cl-] = 1.00M, we have:


Ksp(Hg2Cl2) = [Hg2(2+)]*[Cl-]^2 = 1.01x10^-18