GCem Question

[UnimaasMED]

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During a titration it was determined that 30.00 mL of a 0.1 M Ce4+ solution was required to react completely with 20.00 mL of a 0.150 M Fe2+ solution. Which one of the following reactions occured?

Ce4+ + 3 Fe2+ + H2O --> 3 Fe3+ + CeO- +2H+
2 Ce4+ + Fe2+ -->Fe4+ + 2Ce3+
Ce4+ + Fe2+ -->Fe3+ + 2Ce3+
Ce4+ + 2Fe2+ -->2Fe3+ + 2Ce2+
Ce4+ + 2Fe2+ -->2Fe4+ + Ce2+ + 2e-

 

[rpatel8]

First you need to see how much moles of each are reacting. .1M= x/.03L = .003 mol of Ce4+ and .15M=x/.02L = .003 mol Fe2+. Therefore, you know it is reacting in a 1:1 matter so the correct answer is the 3rd choice Ce4+ + Fe2+ -->Fe3+ + 2Ce3+ because you have 1 mol of both reactants

 

[lippy104]

During a titration it was determined that 30.00 mL of a 0.1 M Ce4+ solution was required to react completely with 20.00 mL of a 0.150 M Fe2+ solution. Which one of the following reactions occured?

Ce4+ + 3 Fe2+ + H2O --> 3 Fe3+ + CeO- +2H+
2 Ce4+ + Fe2+ -->Fe4+ + 2Ce3+
Ce4+ + Fe2+ -->Fe3+ + 2Ce3+
Ce4+ + 2Fe2+ -->2Fe3+ + 2Ce2+
Ce4+ + 2Fe2+ -->2Fe4+ + Ce2+ + 2e-

Its the third one.

First you need to find out the moles of each reacted. 30*0.1 = 3 mmol, and 10*0.15 = 3 mmol. So the stoichiometric coefficients in the reaction should be equal.

 

[UnimaasMED]

thanks 👍

 

[rippinitez]

I'm confused, so how is it equal if there is a 2 in front of Ce3+?

 

[rpatel8]

I'm confused, so how is it equal if there is a 2 in front of Ce3+?

All you are concerned with is what is reacting. There is 1 mol of each of the reactants so the reactants should be 1:1 ratio. Then you balance the products in terms of what you get.