Redox Sample Question 63 e

Redox Sample Question 63 e - 4 2-+ 3IO 4-Next, determine...

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63. e) Balance the following redox reaction in basic solution: CrI 3 + Cl 2 CrO 4 2- + IO 4 - + Cl - First, we must assign oxidation states for all the atoms. Reactants: For CrI 3 ; the oxidation states are Cr = +3, and I = -1 For Cl 2 ; the oxidation states are Cl = 0 Products: For CrO 4 2- ; the oxidation states are Cr = +6, O = -2 For IO 4 - ; the oxidation states are I = +8, O = -2 For Cl - ; the oxidation states is Cl = -1 Now, determine which atoms are being reduced and which are being oxidized. First, we write out the reduction part of the half reaction: Cl 2 + 2e - 2Cl - Now write out the oxidation part of the half reaction. Notice that both Cr and I are oxidized, so we include both of these in balancing the half reaction (this is always true when we have more than one atom species being oxidized or reduced): CrI 3 CrO
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Unformatted text preview: 4 2-+ 3IO 4-Next, determine the amount of electrons we need to add. The difference in the Cr oxidation states are 3 electrons, and the difference between the iodine oxidation states are 8 electrons. Thus, the change in electrons is: 3 + 3(8) = 27 electrons. CrI 3 CrO 4 2-+ 3IO 4- + 27 e-Balance the charges with hydroxide (OH-); 3+2+27= 32: CrI 3 + 32OH- CrO 4 2-+ 3IO 4- + 27 e-Finally, balance the hydrogens with water (H 2 O): CrI 3 + 32OH- CrO 4 2-+ 3IO 4- + 27 e-+ H 2 O Now that we have balanced our half reactions, we need to equate the electrons in the two half reactions. 27Cl 2 + 54e- 54Cl-+ 2CrI 3 + 64OH- 2CrO 4 2-+ 6IO 4- + 54 e-+ 2H 2 O 27Cl 2 + 2CrI 3 + 64OH- 2CrO 4 2-+ 6IO 4- + 54Cl-+ 2H 2 O And now we are done!...
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