answer question 4 show all work Equeations to Balance Balance the redox reaction

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answer question 4 show all work

Equeations to Balance Balance the redox reactions listed using the ion-electron method. Note that the chemical species while the chemical containing the element being reduced is referred to as the oxidizing agent wh pecies containing the element being oxidized is referred to as the reducing agent. For example, in the reaction on the previous page. Mn is reduced and therefore. MnO, is the oxidizing agent oxidized making CIO2 the reducing agent. Identity the oxidation and reduction half reactions, the elements oxidized and reduced, and the oxidizing and reducing agents for each reaction shown ai, in acid I. Cu + NO3- Cu+2 + NO, S-03-2 + 1, s,06-2 + 1. 5. C204.2 + MnO4 CO2 + Mn+2 3. 4. Br2 Br-+ Br03. 7, Zn + NO3 Zn+2 + NH4+ 9, Cr207-2 + CH3CH2OH Cr+3 + CH3CHO 10. Zn + V03 Zn+2 + V+2 11. IO3 I13 12. Pt+N03 + cr PtC16-2 + NO 14. MnO4-+ HS03 Mn+2 + S04-2 5. C104 + HAsO--> H3Aso, + cr H2O2 + Cr-07-2-Cr+3 + O2 Sn+O2 Sn+4 CNS + MnO4 CO2 + NO + SO2 + Mn+2

Explanation / Answer

Answer:

Br2 -------> Br- + BrO3-

Separating this reaction into oxidation and reduction half reations :

reduction half reaction: Br2 ------> Br-.....(1)
oxidation half reaction: Br2--------> BrO3-......(2)

Now balancing reduction half first - balance all atoms other than O and H ....

Br2 -------> 2Br-.....(4)

Now balancing oxidation number by adding electrons....oxidation no. of Br in Br2 is 0 while in Br- is -1 ...thus each Br accpets 1 electron.since there are 2 Br atoms on right side of eqn (4) therefore adding 2 e- to left side of eqn (4)

Br2 + 2e- --------> 2Br- ....(5)

balancing charge - not needed as charge on either side is already balanced ...

thus eqn(5) represents balanced reduction half reaction

now balancing oxidation half reaction...

First balancing all atoms other than O and H....

Br2 ------> 2BrO3- ........(6)

Now balancing oxidation no. by adding electrons....oxidation of Br in Br2 is 0 and in BrO3- is +5 . thus each Br atom looses 5 electrons since there are two atoms of Br in right side of (6) therefore adding 10 e- to right side of (6)..

Br2 --------> 2BrO3- + 10e-.......(7)

Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

O: Br2 + 6H2O 2BrO3- + 10e- + 12H+

R: Br2 + 2e- 2Br-

Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

O:Br2 + 6H2O 2BrO3- + 10e- + 12H+

R: Br2 + 2e- 2Br-

O: Br2 + 6H2O 2BrO3- + 10e- + 12H+

R: 5Br2 + 10e- 10Br-

Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

6Br2 + 6H2O + 10e- 2BrO3- + 10Br- + 10e- + 12H+

Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

3Br2 + 3H2O BrO3- + 5Br-

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