How do you balance #Cr + Fe(NO_3)_2 -> Fe + Cr(NO_3)_3#?

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Answer 1 · 2016-12-28T07:58:47

#2Cr+3Fe(NO_3)_2 rarr 3Fe + 2Cr(NO_3)_3#

We write the oxidation rxn:

#CrrarrCr^(3+) + 3e^-# #(i)#

And then the reduction rxn:

#Fe^(2+) +2e^(-) rarr Fe# #(ii)#

And then cross mulitply, #2xx(i) +3xx(ii)#

#2Cr+3Fe(NO_3)_2 rarr 3Fe + 2Cr(NO_3)_3#

Is mass balanced? Is charge balanced?