At 900 degrees C, Kp = 1.04 for the reaction CaCO3 (s) –><– CaO (s) + CO2 (g). At a low temperature, dry ice (solid CO2),

At 900 degrees C, Kp = 1.04 for the reaction CaCO3 (s) -->

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The answer is d.potassium
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Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======begin{aligned} text{i}_2text{footnotesize (aq)}+2,text{e}^{-} & rightleftharpoons 2,text{i}^{-}text{footnotesize (aq)}& text{e} & = +0.53text{ v} \ text{co}^{3+}text{footnotesize (aq)} + text{e}^{-} & rightleftharpoons text{co}^{2+}text{footnotesize (aq)} &  text{e} & = +1.80text{ v} \ text{o}_2text{footnotesize (g)}+4,text{h}^{+}text{footnotesize (aq)} + 4,text{e}^{-} & rightleftharpoons 2,text{h}_2text{o}text{footnotesize(l)}&  text{e} & = +1.23text{ v} \ end{aligned}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.

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