For the following balanced solution, determine E o forthis reduction half-reacti

Question

For the following balanced solution, determine Eo forthis reduction half-reaction using the following data and a tableof standard reduction potentials.

8H+ + BaCrO4 + 3e-------------> Ba2+ + Cr3+ +4H20

Ksp (BaCrO4) = 2.1x10-10
Substance:                      Gof (kJ/mol)
CrO42-                               -727.75
Cr2O72-                            -1301.1
H+                                       0
H20                                    -237.18

Explanation / Answer

                     BaCrO4      ->         Ba2+  +         CrO42-                      (1)           H+ +  CrO42-        ->   0.5 H2O   +    0.5Cr2O72-                     (2) 3e- + 7H+ +0.5Cr2O72-   ->   3.5 H2O   +         Cr3+                                   (3) -------------------------------------------------------------- 8H+ + BaCrO4 + 3e-------------> Ba2+ + Cr3+ +4H20 delG1 = -RT*lnKsp = -(8.314 J/mol/K) *298K *ln(2.1e-10) =55.21 kJ delG2 = [0.5*(-1301.1) + 0.5*(-237.18)] - [0 + 1*(-727.75)]= -41.39 kJ Third reaction is given by E0 = 1.33 V (look this common reactionup in standard cell potential tables); delG3 = -nFE0 = -3 mol e-*(96485 C/mol e-)*1.33 V = -384.98 kJ where C*V = 1 J overall delG = -371.16 kJ delGoverall = -nFE0', where E0' is the potential of the desiredhalf-reaction E0' = 371.16 x103 J /(3 mol e- *96485 C/mol e-) E0' = 1.28 V

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