When the Hg2+ concentration is 1.07 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.664V. What is the Cr3+ concentration?3Hg2+(aq) + 2Cr(s)→ 3Hg(l) + 2Cr3+(aq)Answer: ___ M The Nernst EquationA non-standard cell or half-cell potential can be calculated using the Nernst Equation:RTE = E° - - In QnFwhereEpotential under non-standard conditionsE° = standard potentialideal gas constantT = kelvin temperature= number of moles of electrons for the reaction as writtenF = charge carried by 1 mol of electronsQ = reaction quotientRIt 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!298.15 KF96,485 J V' mol!RT(8.314 J mol" K-l)(298.15 K)0.0257 V-F96,485 J V' mol!and the Nernst equation with the potentials in volts is:PreviousNext and the Nernst equation with the potentials in volts is:0.0257E = E°In Qnatural logarithmSometimes it is more convenient to use base-10 logarithms, and the substitution of 2.303 log for In is made. (2.303 x 0.0257 = 0.0592)Then, the Nernst equation for base-10 logs at 25 °C is:0.0592E = E°log Qbase-10 logarithmA common student error is to use the wrong kind of logarithm. Be sure, when you choose an equation, to use the correct logarithm.

It is given that Ecell for the given cell is 1.664 V at 25°C and the concentration of Hg2+ is 1.07 M and the cell reaction is shown below, 3Hg2+(aq) + 2Cr(s)→ 3Hg(l) + 2Cr3+(aq)To calculate E° for the given cell, At cathode,3Hg2+ + 6e- → 3Hg ; E∘= 0.86 VAt anode,2Cr → 2Cr3+ + 6e- ; E∘= -0.74 VE∘cell = E∘cathode - E∘anode= 0.86 V - (-0.74 V)= 1.6 VTo calculate the concentration of Cr3+, Ecell = E∘ - 0.05926 log[Cr3+]2[Hg2+]31.664 V = 1.6 - 0.05926 log [Cr3+]2(1.07)30.064 = - 0.00986 log [Cr3+]21.225[Cr3+] = 0.000629 M= 6.29×10-4 M