In basic solution, se2 - and so32 - ions react spontaneously and e o cell = 0.35 v. (a) write the balanced half-reactions for this process. include the states for each reactant and product. (b) if e o sulfite is - 0.57 v, calculate e o selenium. 2se2 - (aq) + 2so32 - (aq) + 3h2o (l) → 2se (s) + 6oh - (aq) + s2o32 - (aq)

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Spud

Chemistry

11 July, 17:15

In basic solution, se2 - and so32 - ions react spontaneously and e o cell = 0.35 v. (a) write the balanced half-reactions for this process. include the states for each reactant and product. (b) if e o sulfite is - 0.57 v, calculate e o selenium. 2se2 - (aq) + 2so32 - (aq) + 3h2o (l) → 2se (s) + 6oh - (aq) + s2o32 - (aq)

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Richard Watts · Answer 1

(a) Write balanced half-reactions for the process:

Oxidation: Se^2 - (aq) → Se (s) + 2e-

Reduction: 2So3^2 - (aq) + 3H2O (l) + 4e - → S2O3^2 - + 6OH - (aq)

(b) If E sulfite is 0.57 V, calculate E selenium:

E anode = E cathode - E cell

= - 0.57 - 0.35

= -.092