Use the data given below to construct a born-haber cycle to determine the heat of formation of kcl. δ h°(kj) k(s) → k(g) 89 k(g) → k+(g) + e- 418 cl2(g) → 2 cl(g) 244 cl(g) + e- → cl-(g) -349 kcl(s) → k+(g) + cl-(g) 717 use the data given below to construct a born-haber cycle to determine the heat of formation of kcl. h°(kj) k(s) → k(g) 89 k(g) → k+(g) + e- 418 cl2(g) → 2 cl(g) 244 cl(g) + e- → cl-(g) -349 kcl(s) → k+(g) + cl-(g) 717 -1119 kj +631 kj +158 kj -997 kj -437 kj

Calcium chloride +gold(1) oxide—>

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Oxidation-reduction reactions (often called "redox" for short) are reactions that involve the transfer of electrons from one species to another. oxidation states, or oxidation numbers, allow chemists to keep track of these electron transfers. in general, one element will lose electrons (oxidation), with the result that it will increase in oxidation number, and another element will gain electrons (reduction), thereby decreasing in oxidation number. the species that is oxidized is called the reducing agent or reductant. the species that is reduced is called the oxidizing agent or oxidant. to sum up: oxidation = increase in oxidation state = loss of electrons = reducing agent reduction = decrease in oxidation state = gain of electrons = oxidizing agent part a which element is oxidized in this reaction? fe2o3+3co→2fe+3co2 enter the elemental symbol. view available hint(s) is oxidized part b which element is reduced in this reaction? 2hcl+2kmno4+3h2c2o4→6co2+2mno2+2kcl +4h2o enter the elemental symbol. view available hint(s) is reduced