Question

In basic solution, Sc^2- and SO_3^2- ions react spontaneously: 2Se^2- (aq) + 2SO_3^2- (aq) + 3H_2 O (l) rightarrow 2Se(s) + 6OH^- (aq) + S_2 0_3^2- (aq) E degree _cell = 0.35 V (a) Write balanced half-reactions for the process. (b) If E degree _sulfite is -0.57 V. calculate E degree _selenium.

Answer 1

a)

Oxidation half reaction:

Se^2- (aq) -> Se(s) + 2e^-

Developing reduction half reaction:

Step 1 (Balance S atoms): 2 SO₃^2- -> S₂O₃^2-

Step 2 (Balance O atoms by adding H₂O): 2 SO₃^2- -> S₂O₃^2- + 3 H₂O

Step 3 (Balance H atoms by adding H⁺): 2 SO₃^2- + 6H⁺ -> S₂O₃^2- + 3 H₂O

Step 4 (Balance charge by adding e^-): 2 SO₃^2- + 6H⁺ + 4e^- -> S₂O₃^2- + 3 H₂O

Step 5 (Add 6OH- to both sides for basic solution): 2SO₃^2- + 6H₂O + 4e^- -> S₂O₃^2- + 3 H₂O + 6 OH^-

Reduction half reaction:

2SO₃^2- + 6H₂O + 4e^- -> S₂O₃^2- + 3H₂O + 6OH^-

Oxidation half reaction:

2Se^2- (aq) -> 2Se(s) + 4e^-

b)

E^o_cell = E^o_red - E^o_ox = 0.35 V

E^o_sulfite = E^o_red = -0.57 V

-0.57 V - E^o_ox = 0.35 V

E^o_ox = -0.92 V

E^o_selenium = E^o_ox = -0.92 V