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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?

User Brent Schooley
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1 Answer

14 votes
14 votes

Answer


SeO_2+4OH^-\rightarrow SeO^(2-)_4+2H_2O+2e^-

Step-by-step 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^-

User Renish Aghera
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