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A solution contains Ce(SO4)3 2- at a concentration of 0.0150 M. It was found tha

ID: 896858 • Letter: A

Question

A solution contains Ce(SO4)3 2- at a concentration of 0.0150 M. It was found that in a titration, 25.00 mL of this solution reacted completely with 23.44 mL of 0.032 M FeSO4 solution. The reaction gave me Fe 3+ as a product in the solution. In this reaction, what is the final oxidation state of Ce? A solution contains Ce(SO4)3 2- at a concentration of 0.0150 M. It was found that in a titration, 25.00 mL of this solution reacted completely with 23.44 mL of 0.032 M FeSO4 solution. The reaction gave me Fe 3+ as a product in the solution. In this reaction, what is the final oxidation state of Ce? A solution contains Ce(SO4)3 2- at a concentration of 0.0150 M. It was found that in a titration, 25.00 mL of this solution reacted completely with 23.44 mL of 0.032 M FeSO4 solution. The reaction gave me Fe 3+ as a product in the solution. In this reaction, what is the final oxidation state of Ce?

Explanation / Answer

Solution :-

Lets first calculate the moles of the each reagent

moles = molarity * volume in liter

moles of Ce(SO4)3^2- = 0.0150 mol per L * 0.025 L = 0.000375 mol

moles of FeSO4 = 0.032 mol per L * 0.02344 L = 0.00075 mol FeSO4

mole ratio = moles of FeSO4 / moles of Ce(SO4)3^2-

                = 0.00075 mol / 0.000375 mol

               = 2

therefore mole ratio of the reaction is 1 mol Ce(SO4)3^2- : 2 mol FeSO4

the oxidation state of the Fe changes from +2 to +3 means its oxidizing

thereofore

initial oxidation state of the Ce in Ce(SO4)3^2-

(1*Ce) + (-2*3) = -2

Ce +(-6) = -2

Ce = -2+6

Ce =+4

now after reacting with the FeSO4

FeSO4 it gives 0.00075 mol electrons

therefore the oxidation state of the Ce in the final solution will change by 0.00075/0.000375 = 2

so the oxidation state of the Ce in the final state is +2

(Ce changes from +4 to +2 means its reduced)

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