uided Solut C Y E Cell Calculationi N/bisc For a particular redox reacti C For A
ID: 538108 • Letter: U
Question
uided Solut C Y E Cell Calculationi N/bisc For a particular redox reacti C For A Particular Redox Rea ns/mod/ibis/view.php?id 3734439 25/2017 11:55 PM O 16.8/206/22/2017 03:28 PM Gradebook Print Calculator -Penodic Table on 36 of 38 A solution containing the following was prepared 0.19MPh2. 1.5 104MPo·5x106 MM 2 0.19 M MnO4, and 0.93 M HNOs For this solution, the following balanced reduction half-reactions and overall net reaction can occur - E-1.690v E -1.507 v A) Determine Ecel,AG, and K for this reaction. Number Number Number B) Calculate the value for the cell potential, Erd, and the free energy, a, for the given consens. Number Number Scroll down for the rest of the question C) Calculate the value of Eces for this system at equiibrium Number O Psevous Ch.ck AnsaerNeri 8 E.,Explanation / Answer
Ans Eo cell = Ecathode - Eanode = 1.690-1.507 = 0.183V
Ecell = Eocell -0.0591 / 10 * log(Pb2+)^2 (H+)^16 * (MnO4-)^2 / (pb4+)^5 (Mn2+)^2
= 0.183 - 0.002955 * log(0.19)^2 (0.93)^16 (0.19)^2 / (1.5*10^-6)^5 (1.5 * 10 ^-6)^2
= 0.183 - 0.002955 log(2.39 * 10^37)
= 0.183 - 0.11
Ecell = 0.073V
delta G - nF Ecell = - 10 * 96500 * 0.073 = 70445 J = 70.4KJ
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