Use standard reduction potentials to calculate the equilibrium constant for the reaction:Hg2+(aq) + 2Cr2+(aq)→ Hg(l) + 2Cr3+(aq)Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm.Equilibrium constant: ______G° for this reaction would be _________ (greater or less ) than zero. Relationship between E°cell and KThe equilibrium constant for a cell reaction can be calculated from the standard cell voltage. The relationship between them is:RTE°.cellIn KnFwherestandard cell potential= ideal gas constantkelvin temperature= number of moles of electrons for the reaction as writtenF = charge carried by 1 mol of electronsK = equilibrium constantE°RTnIt is customary to use the equation in a form where numerical values are substituted for R, T and F at a temperature of 25 °C.ForR =8.314 J mol-'K!T298.15 KF =96,485 J V' mol"1(8.314 J mol K-)(298.15 K)RT= 0.0257 VF96,485 J V-l mol!and the equation with the potential in volts is:0.0257PreviousNextE° cellIn Knatural logarithm R =8.314 J mol"'K!T298.15 KF =96,485 J V-' mol"1(8.314 J mol K-)(298.15 K)RT0.0257 V96,485 J V' molFand the equation with the potential in volts is:0.0257E°cellIn Knatural logarithmSometimes the base-10 logarithm is used and the substitution of 2.303 log for In is made. (2.303 x 0.0257 = 0.0592) Then, the equationfor base-10 logs at 25 °C is:0.0592E° celllog Kbase-10 logarithmnA common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correct logarithm.

The given reaction is: Hg+2 (aq) + 2e-→Hg (l) E0 = 0.855 V2Cr+3 (s) +2 e- →2 Cr+2 (aq) E0 = -0.41 VE0cell = 0.855 V --0.41 V=1.265 V(a) The equilibrium constant of the reaction is as follows: Ecell0=0.0592nlog K1.265 = 0.02572log Klog K =1.265×20.0257log K =98.444K=5.6713 ×1042 ΔG0 = - nFE0 = - 2 × 96485 J/(V.mol) × 1.265 V = -244107.05 J/mol. It is less than zero.