10e 1. Two beakers are joined by a salt bridge. One beaker contains aqueous solu
ID: 1010920 • Letter: 1
Question
10e 1. Two beakers are joined by a salt bridge. One beaker contains aqueous solutions of potassium and manganous chloride, Mncl2. The other beaker contains aqueous solutions of potassium permanganate and chloride. Both solutions are acidic. Given that molar concentrations dichromate H 0.100 M, 0.0500 M, 0.0750M, 0.100 M, the of Mnoi, Mn'. and 1.7 10 respectively, what is the emf of this cell? [Hint: Remember, rather than balancing the equation by the half.reaction method, you may used already balanced half-reactions from Appendix E to get started. This is a very straightforward problem; don't make it harder than it has to be.] (10 pts) Mno4 (aq) cr' (aq) Mn2 (aq) crzo' (unbalanced)Explanation / Answer
As you have given the half reaction and their reduction and oxidation potentials, we can add up both equations to get final equation.
Reduction reaction, MnO4- + 8H+ + 5e- -----------> Mn2+ + 4H2O
Multiply this equation by 6.
6MnO4- + 48H+ + 30e- -----------> 6 Mn2+ + 24H2O___________________equation A
Oxidation reaction, 2Cr3+ + 7H2O -----> Cr2O72- + 14H+ + 6e-
Multiply this equation by 5.
10Cr3+ + 35H2O -----> 5Cr2O72- + 70H+ + 30e-____________________equation B
Add equation A and B, we get,
6MnO4- + 10Cr3+ +11H2O --------> 6Mn2+ + 5Cr2O72- + 22H+
Eocell = Eoox + Eored
= -1.33+1.51 = 0.18 V
At 298 K,
EMF = Eo -0.06/n log([Mn2+]6[Cr2O72-]5[H+]22)/([MnO4-]6[Cr3+]10)
= 0.18 - 0.06/30 log[(0.0500)6(0.0750)5(1.7*10-4)22]/[(0.100)6(0.100)10]
= 0.18 - 0.002 log[1.56*10-8)(2.37*10-6)(1.17*105*10-88)]/[(10-6)(10-10)]
= 0.18 - 0.002 log [4.33*10-81]
= 0.18 + 0.161 = 0.341 V
EMF of the cell = 0.341 V
EMF of cell = 0.341 V
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