How can one balance the following redox reaction in basic medium?

Balance this redox reaction in basic medium:
Zn(s)+CrO4^2-(aq) =Zn(OH)4^2-(aq)+CrO4^-

1 Answer
anor277
Oct 5, 2017

Well chromate ion is likely reduced to chromic hydroxide in a basic medium.....

Explanation:

#CrO_4^(2-) +3e^(-) +4H_2O rarr Cr(OH)_3 +5HO^(-) # #(I)#

Charge and mass are balanced so this is kosher.....

And zinc is likely oxidized to #Zn(OH)_4^(2-)#

#Zn + 4HO^(-) rarr Zn(OH)_4^(2-) + 2e^-# #(II)#

And we takes #2xx(I)+3xx(II)# to give.....

#2CrO_4^(2-) +6e^(-) +8H_2O +3Zn + 12HO^(-) rarr 2Cr(OH)_3 +10HO^(-) +3Zn(OH)_4^(2-) + 6e^-#

Which after cancelling gives.....

#2CrO_4^(2-) +8H_2O +3Zn + 2HO^(-) rarr 2Cr(OH)_3 +3Zn(OH)_4^(2-)#

The which I think is balanced with respect to mass and charge (but do not trust my arithmetic!). All I have done here is to assign oxidation states, and balanced mass and charge.

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