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Odixation Chemistry Problem

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. What is the difference between an oxidizing agent and a reducing agent?

2. When first learning to balance equations, we learned that the number of atoms of each element in the products and reactants must be equivalent. What are some additional factors that must be taken into account when balancing equations for redox reactions?

3. What are half reactions?

4. What two aspects of the half-reaction equations must be balanced?

5. For the equation Ag NO3 - à Ag NO

(Note: This reaction takes place in an acidic solution.)

Step 1: What substance is reduced?

Step 2: What substance is oxidized?

Step 3: What is the half reaction for oxidation?

Step 4: What is the half reaction for reduction?

Step 5: What is the net balanced equation?

Step 6: What is the reduced equation?

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  1. Oxidising agent : reducing themselves (gain of one or more electrons)

    Reducing agent: oxidising themselves (loss of one or more electrons)

    2. In balancing the redox reaction, instead of finding the no. of atoms taking part on both the sides, we have to find the oxidisedone and the reduced one by finding their change in oxidation no. before and after the reaction.

    3.Either Oxidation or reduction reactions are half reactions

    4.Find the oxidising agent and reducing agent and find the oxidation no. before and after the reaction of those.  Then equate the no. of electrons transferred among the two.  Then balance the rest of atoms.

    Zn^0  +  NO3^- +  H^+  gives Zn2+  +  NO2   +  H2O  In this reaction Zn is oxidised (reducing agent has lost 2 electron since oxidation no. increased from 0 to +2) and NO3^- is reduced (oxidising agent where N in NO3^- is +5 state to N in NO2 is +1 state i.e. gain of one electron i.e. oxidation no. of N decreased from +5 to +4)

    therefore the balanced equation is Zn + 2NO3^- + 4H^+ gives Zn^2+  +  2NO2  +  2H2O

    5.(5)  NET BALANCED EQUATION IS 3Ag + NO3^-  +  4H^+  GIVES 3Ag^+  +  NO   +  2H2O  

    NO3^- IS REDUCED

    Ag IS OXIDISED

    Ag TO Ag^+ e^_

    4, 6.) NO3^-  +  3e^-  GIVES NO  

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