A solution of Fe2+ was oxidised to Fe3+ by titration with potassium bromate (KBr

Question

A solution of Fe2+ was oxidised to Fe3+ by titration with potassium bromate (KBrO3). In the reaction BrO3- is reduced to Br-.

(i) Write out a balanced equation for the reaction. (3 marks)

(ii) The concentration of the potassium bromate solution was 0.106 mol dm-3. A 25cm3 aliquot of the iron solution required a titre of 24.5 cm3 of the potassium bromate. Calculate the concentration (in mol dm-3) of the iron solution. .....(4 marks)

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Explanation / Answer

(i) Oxidation half reaction: Fe2+ ------> Fe3+

The balanced oxidation half reaction is: Fe2+ ------> Fe3+ + e-

Reduction half reaction: BrO3- ------> Br-

Balance O on both sides

BrO3- ------> Br- + 3OH-

Balance H by adding H+

BrO3- + 3H+------> Br- + 3OH-

Balance electrons

BrO3- + 3H+ + 6e- ------> Br- + 3OH-

Add H+ on both sides

BrO3- + 6H+ + 6e- ------> Br- + 3H2O

This is balanced half reduction reaction

Add both oxidation and reduction half reactions.

Fe2+ ------> Fe3+ + e- ]*6

6Fe2+ ------> 6Fe3+ + 6e-

BrO3- + 6H+ + 6e- ------> Br- + 3H2O

We get:

6Fe2+ + BrO3- + 6H+   ------> 6Fe3+ + Br- + 3H2O

This is the balanced equation of the reaction.

(ii) [KBrO3] = 0.106 mol/dm3 = 0.106 M

Volume used = 24.5 ml = 0.0245 L

Moles of KBrO3 = concentration*volume = 0.106*0.0245 = 0.0026 mol

0.0026 mol of KBrO3 are used.

In reaction 1 mol of KBrO3 uses 6 moles of Fe2+.

So, 0.0026 mol of KBrO3 uses = 6*0.0026 = 0.0156 mol

Concentration of Fe2+ = moles/volume = 0.0156/0.0245 = 0.636 M = 0.636 mol/dm3

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