Pre-Laboratory Assignment 16 Name: Section Date: Survey of Oxidation and Reducti

Question

Pre-Laboratory Assignment 16 Name: Section Date: Survey of Oxidation and Reduction Reactions 1. Identify the oxidations Cl Cl- HOCI Cio2 CIo 2. Potassium iodate (KIO,) is a strong oxidizing agent and will oxidize Fe2+ ions to Fe3+. In doing so, iodate ion TO,) is reduced to elemental iodine (2). a. Use the oxidation state rules (Gse the Background section) to assign oxidation states to the iodine atoms in iodate ion (10, and iodine (I2). b. The half-reaction for the reduction of iodate is shown below. Use the difference in oxidation states for the iodine atoms in IO3 and I2 to determine the number of electrons gained in this half-reaction. Hint: Hydrogen ions (HH) and water molecules (H,O) are required to balance mass and charge. 3. Combine the oxidation half-reaction for Fe? (see the Background section) with the reduction half-reaction for iodate (Question 2b) and write the balanced equation for the overall redox reaction of Fe+ with IO. Hint: The number of electrons on each side must cancel out.

Explanation / Answer

1. a)The Oxidation number of a pure gas in elemental form is 0 unless it is combined with other atom or exists in ionic form. So, Cl2 has an oxidation number of 0.

b) Cl- exists in ionic form. and so, the overall charge will be the number of ions present. So oxidation number is -1

c) HOCl: The overall oxidation number should be zero. For this to be satisfied, The sum of charges on H,Cl,O should be equal to zero. H exists in +1 state, while O exists in -2 state

Let us assume the charge on Cl = x

Then,             1-2+x=0

                         x= +1

So, Oxidation number of Cl in HOCl = +1

d) ClO2 : This is similar to the previous case. Let us assume the charge on Cl = x

       O exists in -2 state

                                      SO, x+(-2*2) = 0 ;Since two atoms of O are present, oxidation number of O is multiplied with 2   On solving, x=+4

e) ClO3- : In this case, the overall charge = -1

     So, Let us assume the charge on Cl = x

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

On solving, x= +5

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