Question

Consider the following half-reaction balanced for an acidic solution: 2H2O + SeO2 → SeO42- + 4H+ + 2e-. What is the balanced half-reaction for a basic solution?

Answer 1

Answer

$SeO_2+4OH^-\rightarrow SeO^{2-}_4+2H_2O+2e^-$

Explanation

The given balanced half-reaction for an acidic solution:

$2H_2O+SeO_2\rightarrow SeO^{2-}_4+4H^++2e^-$

What to find:

Tha balanced half-reaction for a basic solution.

Step-by-step-solution:

To balance the half-reaction for a basic solution;

1. Add OH⁻ ions to BOTH SIDES to neutralize any H⁺

$2H_2O+SeO_2+4OH^-\rightarrow SeO^{2-}_4+4H^++4OH^-+2e^-$

2. Combine H+ and OH- to make H2O.

$2H_2O+SeO_2+4OH^-\rightarrow SeO^{2-}_4+4H_2O+2e^-$

3. Simplify by canceling out excess H2O

$SeO_2+4OH^-\rightarrow SeO^{2-}_4+2H_2O+2e^-$

4. Balance the charges by adding e-

$SeO_2+4OH^-\rightarrow SeO^{2-}_4+2H_2O+2e^-$