Balance each of the following redox reactions occurring in acidic solution.?

1. I ^- (aq) + NO2 ^- (aq) ---> I2(s) + NO2(g)

2. IO3^-(aq) + H2SO3(aq) ---> I2 (aq) + SO4 ^2-(aq)

3. MnO2(s) + Cu(s) ----> Cu ^2+ (aq) + Mn ^2+ (aq)

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  • 10 years ago
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    Method to follow when balancing these types of equations:

    1. Check that all the non-oxygen atoms are balanced.

    I^- (aq) + NO2^- (aq) ---> I2(s) + NO2 (g)

    Here Iodine is not balanced, there is 1 I^- on the LHS and 2 atoms of I on the RHS.

    Need to make them equal so:

    2I^- (aq) + NO2^- (aq) ---> I2(s) + NO2 (g)

    2. Check that oxygen atoms are balanced on both sides. If they are then leave it as it is. If they are not then we need to add H2O to the side with the least number of oxygen. In this case, we are okay.

    2I^- (aq) + NO2^- (aq) ---> I2(s) + NO2 (g) Oxygens are balanced

    3. Check that the electrons or charges are balanced on both sides. If they are not then we need to add electrons to the side that is not balanced.

    2I^- (aq) + NO2^- (aq) ---> I2(s) + NO2 (g)

    In this case, we have a total of 3 electrons on the LHS and none on the RHS, that is, 2I^- = 2 electrons plus NO2^- = 1 electron, thus we need to add 3 electrons on the RHS to balance that out, hence:

    2I^- (aq) + NO2^- (aq) ---> I2(s) + NO2 (g) + 3e

    Using the same steps we can balance the second redox reaction:

    Step 1: Balance atoms other than oxygen:

    2IO3^-(aq) + H2SO3(aq) ---> I2 (aq) + SO4 ^2-(aq) (Iodine needs to be balanced)

    Step 2: Balance oxygens by adding H^+ and H2O. Add H2O to the side with lowest number of O's:

    2IO3^-(aq) + H2SO3(aq) ---> I2 (aq) + SO4 ^2-(aq)

    (LHS has 6 O's and RHS has 4 O's, so H2O goes into the RHS. Adjust the moles of H2O so that total O and H on RHS equals LHS)

    2IO3^-(aq) + H2SO3(aq) + 8H^+ (aq) ---> I2 (aq) + SO4 ^2-(aq) + 5H2O

    Step 3: Balance the electrons or charges

    2IO3^-(aq) + H2SO3(aq) + 8H^+ (aq) + 8e ---> I2 (aq) + SO4 ^2-(aq) + 5H2O

    (LHS: 2- + 8+ = +6 and RHS: 2- : we need to reduce the electrons on the LHS to -2 by adding 8e, since +6-8 = -2)

    Our final balanced equation is:

    2IO3^-(aq) + H2SO3(aq) + 8H^+ (aq) + 8e ---> I2 (aq) + SO4 ^2-(aq) + 5H2O

    The last equation is done in a similar way:

    MnO2(s) + Cu(s) ----> Cu ^2+ (aq) + Mn ^2+ (aq)

    MnO2(s) + Cu(s) + 4H^+ (aq)----> Cu ^2+ (aq) + Mn ^2+ (aq) + 2H2O

    (2H2O because the RHS has 2 O's and 4H^+ on the LHS to balance the H in 2H2O)

    MnO2(s) + Cu(s) + 4H^+ (aq)----> Cu ^2+ (aq) + Mn ^2+ (aq) + 2H2O

    (No need to add electrons since charges are balanced on both sides)

    Source(s): Myself. I am a Chemistry teacher.
  • Anonymous
    10 years ago

    it is a little hard to understand... but here goes.

    (giving straight answers - without state symbols)

    1. 2I^- + NO2^- ---> I2 + NO2 + 3e^-

    2. 10e^- + 10H^+ + 2IO3^- + H2SO3 ---> I2 + SO4^2- + 5H2O

    3. 4H^+ + MnO2 + Cu ----> Cu^2+ + Mn^2+ + 2H2O

    -add Waters (H2O) to balance Oxygens

    -add Hydrogen ions (H^+) to balance Hydrogens

    -add electrons to balance charges

    Hope this helped.

    Source(s): AS Chemistry.
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  • 3 years ago

    a million. Ni + Cd2+ = Ni2+ + Cd 2. Zn + 2Na+ = 2Na + Zn 2+ 3. 3MnO4- + 24H+ + 5Al = 3Mn2+ + 12H2O + 5Al3+ steps a million. write the equation 2. stability oxigen including H2O 3. balnce H including H+ 4. stability the cost including electrons

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  • 4 years ago

    Number 2 should be:

    2IO3−(aq)+5H2SO3(aq)→I2(aq)+8H+(aq)+5SO42−(aq)+H2O(l)

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