Write a balanced oxidation reduction equation in acidic solution for: #C_2O_4^(2-)+Cr_2O_7^(2-)\toCO_2+Cr^(3+)#?

2 Answers
Jul 14, 2018

AS you have twigged this is a redox reaction, which we can solve by the method of half equations...

Explanation:

#"Oxalate ion"#, #C_2O_4^(2-)#, i.e. #C(+III)# is completely oxidized up to carbon dioxide #C(+IV)#;

#C_2O_4^(2-) rarr 2CO_2(g)+2e^(-)# #(i)#

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

And meanwhile dichromate, #Cr(+VI)# is reduced to #Cr(+III)#..

#Cr_2O_7^(2-)+14H^+ +6e^(-) rarr 2Cr^(3+) + 7H_2O# #(ii)#

We take #3xx(i)+(ii)# to retire the electrons, virtual particles of convenience, to give....

#underbrace(Cr_2O_7^(2-))_"orange-red" +3C_2O_4^(2-)+14H^+ rarr underbrace(2Cr^(3+))_"green"+6CO_2(g) + 7H_2O#

….after cancellation...

And if this AIN'T balanced with respect to mass and charge, then it cannot be accepted as a representation of chemical reality .. Is it right?

We can usually assign oxidation numbers to organic compounds give the same rules that we employ in inorganic chemistry. Methane is #C(-IV)#...#CO-=C(+II)#...ethane is #C(-III)#...#CO_2-=C(+IV)#...and #"propane"# #-=# #H_3stackrel(-III)C-stackrel(-II)CH_2CH_3#

Jul 14, 2018

Please see the equation below

Explanation:

The half-reactions are

#Cr_2O_7^(2-)+14H^+ + 6e^(-) rarr 2Cr^(3+)+7H_2O#

#C_2O_4^(2-)rarr 2CO_2+2e^-#

Therefore,

#Cr_2O_7^(2-)+14H^+ + 6e^(-) rarr 2Cr^(3+)+7H_2O#

#3C_2O_4^(2-)rarr 6CO_2+6e^-#

Adding the last #2# equations

#Cr_2O_7^(2-)+14H^+ +3C_2O_4^(2-) rarr 2Cr^(3+)+6CO_2+7H_2O#