Question

balance the following redox reaction in basic conditions:

BiO2- + Te4O9 → Bi82+ + Te(OH)6

Answer 1

BiO2- + Te4O9 → Bi82+ + Te(OH)6

Split in half reactions

BiO2- → Bi8+2

Te4O9  → Te(OH)6

Balance atoms other than O, H

8BiO2- → Bi8+2

Te4O9  → 4Te(OH)6

Balance Oxygen, adding H2O

8BiO2- → Bi8+2 + 16H2O

15H2O + Te4O9  → 4Te(OH)6

Balance Hydrogen adding H+ ions

32H+ + 8BiO2- → Bi8+2 + 16H2O

15H2O + Te4O9  → 4Te(OH)6 + 6H+

Balance charges with electrons

32H+ + 8BiO2- → Bi8+2 + 16H2O

(32 + 8*-2) = +2 + 0

+16 = +2

You need to add 14 electrons in the left

14e- + 32H+ + 8BiO2- → Bi8+2 + 16H2O

For

15H2O + Te4O9  → 4Te(OH)6 + 6H+

0 + 0 = 0 + 6

you need to add 6 electrons in the right

15H2O + Te4O9  → 4Te(OH)6 + 6H+ + 6e-

Now

14e- + 32H+ + 8BiO2- → Bi8+2 + 16H2O

15H2O + Te4O9  → 4Te(OH)6 + 6H+ + 6e-

Balance electrons (14 and 6)

3(14e- + 32H+ + 8BiO2- → Bi8+2 + 16H2O)

7(15H2O + Te4O9  → 4Te(OH)6 + 6H+ + 6e-)

You get

42e- + 96H+ + 24BiO2- --> 3Bi8+2 + 48H2O

105H2O + 7Te4O9 --> 28Te(OH)6 + 42H+ + 42e-

Add both equations (e- should cancel each other)

42e- + 96H+ + 24BiO2- + 105H2O + 7Te4O9 --> 3Bi8+2 + 48H2O + 28Te(OH)6 + 42H+ + 42e-

Cancel common terms

96H+ + 24BiO2- + 105H2O + 7Te4O9 --> 3Bi8+2 + 48H2O + 28Te(OH)6 + 42H+

54H+ + 24BiO2- + 58H2O + 7Te4O9 --> 3Bi8+2 + 28Te(OH)6

This is acidic solution, we require basic so add OH- ions to cancel H+ and form H2O

54OH- + 54H+ + 24BiO2- + 58H2O + 7Te4O9 --> 3Bi8+2 + 28Te(OH)6 + 54OH-

Cancel OH- + H+ = H2O

This is the balanced equation

54H2O + 24BiO2- + 58H2O + 7Te4O9 --> 3Bi8+2 + 28Te(OH)6 + 54OH-

Check balance of H+ ions, H2O, atomic species and electrons