complete the equation and balance Experiment 8 Name: oxidation Reduction: Writin

Question

complete the equation and balance

Experiment 8 Name: oxidation Reduction: Writing Equations Objectives: 1. To identify products in oxidation-reduction reactions 2. To identify oxidizing agents and reducing agents 3. To balance oxidation-reduction reactions by the half-reaction me Introduction: In this experiment we will use the techniques and tables developed in Experiment 7 to identify oxidation and reduction products when various reagents are mixed. Once the products have been identified, the redox can be balanced by the half-reaction method. Notice that the equation must be balanced in acidic conditions if acid has added to the re The equation must be balanced in basic conditions if the reaction has been conducted after adding sodium hydroxide Safety: GOGGLES MUST BE woRN FOR THE Potassium permanganate, potassium chromate, and potassium dichromate are strong oxidizing agents that can cause severe irritation to skin and eyes. Chromates may be carcinogenic Potassium permanganate will skin Sulfuric acid and sodium hydroxide are severely corrosive to eyes and skin. Avoid contact, and wash thoroughly with water after any contact Make sure you dispose of all waste as indicated. Solutions containing any form of chromium or manganese must ver be disposed of in the sink! Resources See handout: Balancing Redox Equations by the Half-Reaction Method and/or see the chapter in the textbook about electrochemistry (normally chapter 2 Procedure Work alone on all parts of the experiment (and report) We will assume that one "squirt" is equal to 1 mL (more precise measurements are not required) the second reagent (with careful mixing) until the color change is complete. Add 3 ml of deionized water to 1 mL of potassium permanganate. In the hood, add sodium sulfide drop by drop until color has disappeared. The oxidation product is elemental sulfur (S). The color of the manganese the perman The reduction product is Mn product is formula (color) Complete the equation and balance it by the half-reaction method: (is it acidic, basic, or neutral? i.e., which solutions combined?) Mno4

Explanation / Answer

Answer 1:

When combining KMnO4 in water with sodium sulfide..

Reduction half reaction: 3 e + 2 H2O + MnO4 MnO2 + 4 OH

Oxidation half reaction: 2 OH + S2 S(e) + H2O + 2 e

combining the half reactions and balancing for electron first

2KMnO4 + 3Na2S + 4H2O = 2MnO2 + 3S + 2KOH + 6NaOH

As Na2S is basic and products are also strong base the reaction is in basic medium.

Answer 2:

Fe2+ + MnO4 - = Fe3+ + Mn2+

Write the oxidation and reduction half-reactions.

Fe2+ = Fe3+ + e- (Charge balanced)

5e- + MnO4-= Mn2+ (Charge imbalanced)

Balance the charges (with H+ because this is in acidic solution) 5e- + 8H+ + MnO4 -= Mn2+ (Charge balanced)

Balance O with H2O

5e- + 8H+ + MnO4-= Mn2+ + 4H2O Hydrogen is Balanced

Add half reactions after cross multiplying by the number of electrons in each 5x (Fe2+ =Fe3+ + e- )

5e- + 8H+ + MnO4-= Mn2+ + 4H2O

5e- + 8H+ + MnO4- + 5Fe2+ = Mn2+ + 5Fe3+ + 5e- + 4H2O

Cancel those species that remain unchanged (in this case only e- , but often we will have H2O, H+ , or OH- on both sides)

8H+ + MnO4- + 5Fe2+ = Mn2+ + 5Fe3+ + 4H2O this is a suitably balanced net ionic equation

Add back the K+ and SO42-

4H2SO4 + KMnO4 + 5FeSO4 = MnSO4 +5/2 Fe2(SO4)3 + K+ + 1/2SO42-+ 4H2O

Multiply by 2 to eliminate the fractional coefficients

8H2SO4 +2 KMnO4 + 10FeSO4 = 2MnSO4 + 5Fe2(SO4)3 + K2SO4+ 8H2O

Answer 3:

let us write

Oxidation half reaction: Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O

The reduction half reaction is : Fe2+ Fe3+ + e-  

and the total balanced reaction is:

Cr2O72- + 6 Fe2+ + 14H+ 2Cr3+ + 6 Fe3+ + 7H2O

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