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)

(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

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.