What is the cofficient of Cr^2+?

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1 Answer
Apr 11, 2018



Start by finding what has been oxidized and what has bee reduced by inspecting the oxidation numbers:

In this case:
#Cr^(2+)(aq) -> Cr^(3+)(aq)#
Is the oxidation
#SO_4^(2-)(aq) -> H_2SO_3(aq)#
is the reduction
Start by balancing the half equations for oxygen by adding water:

#SO_4^(2-)(aq) -> H_2SO_3(aq) + H_2O(l)#
(Only the reduction includes oxygen)

Now balance hydrogen by adding protons:

#4H^(+)(aq)+SO_4^(2-)(aq) ->H_2SO_3(aq) + H_2O(l)#
(again, only the reduction involves hydrogen)

Now balance each half-equation for charge by adding electrons to the more positive side:
#Cr^(2+) -> Cr^(3+) + e^-#

#4H^(+)+SO_4^(2-)(aq)+2e^- -> H_2SO_3(aq) + H_2O(l)#

And to equalize the electrons, multiply the whole half equation with the least electrons by an integer to equal the other half-equation in the number of electrons, thereby balancing the electrons on both sides of the equation:

#2Cr^(2+) -> 2Cr^(3+) + 2e^-#

Now combine everything and remove the electrons (as they in equal amounts on both sides they can be canceled in this step - otherwise just simplify as far as possible)

#4H^(+)(aq)+SO_4^(2-)(aq)+2Cr^(2+)(aq) -> 2Cr^(3+)(aq)+H_2SO_3(aq) + H_2O(l)#

Now the equation is balanced and we can see that the coefficient to #Cr^(2+)# is 2.