Question #dd295

↳Redirected from "What are valence electrons?"
1 Answer
Doc048
Dec 19, 2017

#5Cr^o(s) + 3MnO_4^-color(white)(')(aq) + 24H^+(aq) rightleftharpoons 5Cr^(+3)(aq) + 3Mn^(+2)(aq) + 12H_2O(l)#

#E^o(cell)#= #E^(o)#(Reduction) - #E^o#(Oxidation) = 1.51v - (-0.74v) = 2.25v

Explanation:

Generally, the flow of current is from the cell with the more negative cell potential (Oxidation Rxn) to the cell with the more positive cell potential (Reduction Rxn). In this case

#Cr^o(s) rightleftharpoons Cr^(+3) + 3e^-# => Oxidation Rxn
#MnO_4^(-)(aq) + 8H^+(aq) + 5e^(-) rightleftharpoons Mn^(+2)(aq) + 4H_2O(l)# => Reduction Reaction

To balance charge transfer, multiply the oxidation rxn by 5 and the reduction rxn by 3 and add reactions =>

#5Cr^o(s) rightleftharpoons 5Cr^(+3) + 15e^-#
#3MnO_4^(-)(aq) + 24H^+(aq) + 15e^(-) rightleftharpoons 3Mn^(+2)(aq) + 12H_2O(l)#

Net Cell Rxn:
#5Cr^o(s) + 3MnO_4^(-)(aq) + 24H^+(aq) rightleftharpoons5Cr^(+3)(aq) + 3Mn^(+2)(aq) + 12H_2O(l)#

#E^o(cell)#= #E^(o)#(Reduction) - #E^o#(Oxidation) = 1.51v - (-0.74v) = 2.25v

Related questions
Impact of this question
994 views around the world
You can reuse this answer
Creative Commons License