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In basic solution, Se²⁻ and SO₃²⁻ ions react spontaneously:Se²⁻(aq) + 2SO₃²⁻(aq) + 3H₂O(l) \longrightarrow 2Se(s) + 6[OH]⁻(aq) + S₂O₃²⁻(aq) Ecell° = 0.35V (a) Write balanced half-reactions for the process.

Respuesta :

The balanced half cell reaction for Se²⁻ is

[tex]2Se^{2-}(aq) \longrightarrow 2Se(s) + 4e^-[/tex]

For SO₃²⁻,

[tex]2SO_3^{2-}(aq) + 3H_2O(l) \longrightarrow S_2O_3^{2-}(aq) + 6[OH](aq)[/tex]

What are half-cell reactions?

In half-cell reactions, either an oxidation reaction takes place, in which electrons are lost, or a reduction reaction, in which electrons are gained.

The reactions take place in an electrochemical cell where the electrons are consumed at the cathode during reduction and lost at the anode during oxidation.

In galvanic and voltaic cells, in which electrons move from the anode to the cathode through an electrolyte to create an electromotive force EMF, half-cell reactions are useful for them. Oxidation-reduction half-cell reactions can also happen on metal surfaces and can result in corrosion.

To write the balance half cell reactions

  • Divide the overall reaction into half-reaction

[tex]Se^{2-}(aq) \longrightarrow Se(s) + 2e^-[/tex]

[tex]SO_3^{2-}\longrightarrow S_2O_3^{2-}[/tex]

For Se, no other Oxygen and a hydrogen atom is needed so, balance the charge on both sides

[tex]Se^{2-} \longrightarrow Se + 2e-[/tex]

Multiply by 2 to balance electrons

[tex]2Se^{2-} \longrightarrow 2Se + 4e-[/tex]

For SO₃²⁻,

  • Balance S atoms on both sides

[tex]2SO_3^{2-}\longrightarrow S_2O_3^{2-}[/tex]

  • Balance O atoms with H2O

[tex]2SO_3^{2-}\longrightarrow S_2O_3^{2-} + 3H_2O[/tex]

  • Balance H atoms using H+

[tex]2SO_3^{2-}+ 6H^+\longrightarrow S_2O_3^{2-} + 3H_2O[/tex]

  • Balance the charge on both sides by adding e

[tex]2SO_3^{2-}+ 6H^+ + 4e^- \longrightarrow S_2O_3^{2-} + 3H_2O[/tex]

  • Add 6 OH- both sides

[tex]2SO_3^{2-}+ 6H^+ + 4e^- + 6OH^- \longrightarrow S_2O_3^{2-} + 3H_2O + 6OH^-[/tex][tex]2SO_3^{2-}+ 6H_2O + 4e^- \longrightarrow S_2O_3^{2-} + 3H_2O +6OH^-[/tex]

  • Now balance out the reaction

[tex]2SO_3^{2-}+ 6H_2O + 4e^- \longrightarrow S_2O_3^{2-} +6OH^-[/tex]

Hence, the balanced half-reaction is

Oxidation: [tex]2Se^{2-}(aq) \longrightarrow 2Se(s) + 4e^-[/tex]

Reduction:[tex]2SO_3^{2-}(aq) + 3H_2O(l) \longrightarrow S_2O_3^{2-}(aq) + 6[OH](aq)[/tex]

Learn more about half-reactions:

https://brainly.com/question/2491738

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