Balance the following redox reaction in basic solution. TeO₃²⁻(aq) + As(s) → Te(s) + H₂AsO₄⁻(aq)

Question

Balance the following redox reaction in basic solution

Complete and balance the following redox reaction in basic solution. Be sure to include the proper phases for all species within the reaction.

TeO₃²⁻(aq) + As(s) → Te(s) + H₂AsO₄⁻(aq)

The easiest way to balance this redox reaction is to use the half-reaction method and initially balance it as if it were in acidic solution. Then, once that is done, convert it to a basic solution.

Assigning oxidation numbers:

+4 0 0 +5

TeO3(2-) (aq) + As(s) ------> Te (s) + H2AsO4(-)

So, the As was oxidized (oxidation number increased) and

the Te was reduced (oxidation number was reduced).

Remember oxidation is the loss of electrons and reduction is the gain of electrons.

(OX): As(s) + 4H2O(l) ------> H2AsO4(-)(aq) + 6H(+) (aq) + 5e(-)

The 5e(-) were shown as lost and were necessary to balance charge.

(RED): TeO3(2-) (aq) + 6H(+) + 4e(-) ------> Te (s) + 3H2O (l)

The 4e(-) added show the gain of electrons and to balance the charge.

Since electrons are conserved (Law of conservation of mass), there needs to be the same number of electrons lost as gained. Finding a common denominator, you can multiply the oxidation half reaction by4 and the reduction reaction by 5.

(OX): 4 ( As(s) + 4H2O(l) ------> H2AsO4(-)(aq) + 6H(+) (aq) + 5e(-) )

(RED): 5 (TeO3(2-) (aq) + 6H(+) + 4e(-) ------> Te (s) + 3H2O (l) )

Analyzing both sides of the equations after multiplying the first equation by 4 and the second by 5, one can remove 24 H(+) from both sides and 15 H2O(l) from both sides, leaving

5TeO3(2-)(aq)+6H(+) +4 As(s) +H2O(l) ------> 5Te (s) + 4H2AsO4(-)

This is now balanced, but it shows that the solution requires acidic H(+) ions to be added. Yet the solution is basic. To correct this, add equal number of moles of OH(-) as H(+) to each side.

5TeO3(2-)+6H(+)+4As(s)+H2O+6OH(-) ------>5Te(s)+ 4H2AsO4(-)+6OH(-)

6OH(-) + 6H(+) yield 6H2O, so the final balanced redox reaction in basic solution is:

5TeO3(2-)(aq) +4As(s) +7H2O(l) ------> 5Te(s) +4H2AsO4(-)(aq) +6OH(-)(aq)

Answer 1

5TeO₃^2-(aq) + 4As(s) +7H₂O(l) ==> 5Te(s) + 4H₂AsO₄^-(aq) + 6OH^-(aq) .. balanced redox equation