Balance the redox reaction by inserting the appropriate coefficients

Question

Balance the redox reaction by inserting the appropriate coefficients

redox reaction: H₂O+Se+Cd²⁺--->Cd+SeO₃^2-+H⁺

Answer 1

Hi Kylie,

Redox reactions involve the exchange of electrons between two reactants, where one reactant gets "reduced" (gains electrons) and another reactant gets "oxidized," (loses electrons.) Redox reactions are most easily balanced by splitting them into two "half reactions," balancing the half reactions, and then adding them together to get the complete balanced equation.

In this case it's most obvious that Cd is gaining electrons because it goes from being +2 to neutral. In order to do that it must have gained two electrons.

Cd²⁺ + 2e^- ---> Cd

The rest of the reactants must be involved in oxidation. Se starts out as a neutral atom with oxidation number equal to 0. In the products, the oxidation state of the Se has changed. It is now combined with 3 oxygen atoms, which carry a -2 charge each. -2 * 3 oxygens = 6 negative charges total for every Se. SeO₃^2- has a net -2 charge. The oxidation state of the Se must be +4. (+4 from Se, -6 from 3 Os, give a net -2 charge overall).

H₂O + Se ---> SeO₃ ^2- + H⁺ + e^-

The above reaction isn't balanced yet. There are 3 Os on the right and only one on the left, so put a 3 in front of H₂O, to balance the Os. That gives 6 Hs on the left, so put a 6 in front of the H⁺.

3H₂O + Se ---> SeO₃ ^2- + 6H⁺ + e^-

Good! BUT the net charge on the right is out of balance. The left is neutral (0 charge) so we need 4e^- on the right to get to neutral.

3H₂O + Se ---> SeO₃ ^2- + 6H⁺ + 4e^-

Combine the Half Reactions:

Cd²⁺ + 2e^- ---> Cd

3H₂O + Se ---> SeO₃ ^2- + 6H⁺ + 4e^-

These two half reactions can't quite be combined until the e^- on both sides are equal and therefore exchange in balance. To do that, multiply the Cd half reaction by two, so it will have 4e^-.

2Cd²⁺ + 4e^- ---> 2Cd

3H₂O + Se ---> SeO₃ ^2- + 6H⁺ + 4e^-

Now add the reactions. The e^- on both sides are the same, so when combined you get the final, balanced redox reaction:

3H₂O + Se + 2Cd²⁺ ---> 2Cd + SeO₃ ^2- + 6H⁺

Always good to check:

H: 6 on left and 6 on right

O: 3 on left and 3 on right

Se: 1 on left and 1 on right

Cd: 1 on left and 1 on right

Charge on left = +4; Charge on = +4.

Yahtzee!