How do we represent the oxidation of #"bismuth metal"# by the action of #"nitric acid"# to give #"bismuth nitrate"#, and #NO(g)#?

1 Answer
Aug 25, 2017

Answer:

You are oxidizing #"bismuth metal"# to #"bismuth(III) nitrate"#.......

Explanation:

Bismuth metal is oxidized to #Bi^(3+)#:

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

Nitrate, #stackrel(+V)NO_3^-#, is reduced to #stackrel(+II)NO#:

#NO_3^(-) + 4H^(+) + 3e^(-) rarrNO + 2H_2O # #(ii)#

Charge and mass are balanced in both equations, and we take #(i) + (ii)#

#Bi+ NO_3^(-) + 4H^(+) rarr Bi^(3+) + NO + 2H_2O #

Is this balanced?

Else....we add #3xxNO_3^-# to each side to get......

#Bi + 4HNO_3 rarr3Bi(NO_3)_3 +NOuarr+2H_2O#