Balancing a RedOx equation in BASIC solution?

Question

Compose the balanced reaction equation for the following reaction occuring in basic aqueous medium.

Reduce all coefficients to the lowest possible integral values.

Include H2O(l) and OH−(aq) appropriately to complete the equation.

Fe(CN)6^4−(aq) Ce^4 (aq) --> Ce(OH)3(s) Fe(OH)3(s) CO3^2−(aq) NO3^−(aq)

I understand how to do this under acidic conditions, but the basic has completely confused me! If some one could aslo explain this process too.

Answer 1

Odd-looking reaction.  I'm not sure how CN- can be oxidized all the way to CO3^2- and NO3- with only Ce^4+ as the oxidizing agent.  Magic, I guess.  But I guess the problem is to balance, not to figure out if the reaction can really happen.

Balancing in basic solution just means you can add OH- and H2O as needed.  As with any redox reaction, balance electrons first, then all speices other than H and O.  Then finally balance H and O.  If you need to add an H to one side of the equation, add H2O that side and OH- to the other.  If you need to add O to one side of the equation, add 2OH- to that side and H2O to the other.  So here goes:

The starting reaction is:

Fe(CN)6^4- (aq) + Ce^4+ (aq) --> Ce(OH)3 (s) + Fe(OH)3 (s) + CO3^2- (aq) + NO3^- (aq)

Balance electrons first:

C goes from 2+ to 4+, N goes from 3- to 5+, Fe goes from 2+ to 3+, and Ce goes from 4+ to 3+.  Ce is the only oxidizing agent, so it has to handle 11 electrons worth:

Fe(CN)6^4- (aq) + 11Ce^4+ (aq) --> 11Ce(OH)3 (s) + Fe(OH)3 (s) + CO3^2- (aq) + NO3^- (aq)

Next, balance OH-, carbon, and nitrogen:

Fe(CN)6^4- (aq) + 11Ce^4+ (aq) + 36OH- (aq) --> 11Ce(OH)3 (s) + Fe(OH)3 (s) + 6CO3^2- (aq) + 6NO3^- (aq)

Finally, balance O and H:

H is balanced.  You need 36 O on the left to reach balance:

Fe(CN)6^4- (aq) + 11Ce^4+ (aq) + 108OH- (aq) --> 11Ce(OH)3 (s) + Fe(OH)3 (s) + 6CO3^2- (aq) + 6NO3^- (aq) + 36H2O (l)

There are no species that are identical on both sides, so you're done.