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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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meerkat18

(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

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