A3.35cm air bubble forms in a deep lake at a depth where the temperature is 19.2 degrees celcuis and the pressure is 1564mmhg. the bubble rises to a depth where the temperatuer is now 26.5 degrees celsius and the pressure is 825.5 mmhg. assuming the amount of air in the buble has not changed, calculate its new volume

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

The answer is solid copper 

dndnf

2poclâ‚(g) ⇜ 2 pclâ‚(g) + oâ‚‚(g) â†gâ° = (2 mol)î”gfâ°(pclâ‚) – (2 mol)î”gfâ°(poclâ‚) = (2 mol)(–270 –(–502) kj/mol) â†gâ° = + 464 kj = â†hⰠ– tî”sâ° î”sâ° = 179 j/k ⇒ â†hâ° = â†gâ° + (298 k)(179 j/k) = –411 kj assuming that î”hâ° and î”sâ° do not depend on temperature, the temperature at which â†gâ° = 0 is where t = â†hâ°/î”sâ° = (–411 kj)/(0.179 j/k) = 2294 k. at that temp. keq = 1 for the reaction.