I am so lost. Please show step by step, Thank you :) 1. In basic solution, Br2 d

Question

I am so lost. Please show step by step, Thank you :)

1. In basic solution, Br2 disproportionates to bromide ions and bromate ions. Use the half reaction method to balance the equation for this reaction: Br2 (l) ---> Br- (aq) + BrO3 (aq)

2. Permanganate ion, MnO4- is used in the laboratory as an oxidizing agent; thiosulfate ion, S2O3^-2 is used as a reducing agent. Write the balanced equation for the reaction of these ions in an acidic aqueous solution to produce manganese (II) ion and sulfate ion.

Explanation / Answer

1)

To balance this redox reaction, we must split this reaction into two half reactions about a common reactant, Br2(l). From the reaction, we see that Br(l) forms two products individually:

Br2(l) ----> BrO3-(aq) AND
Br2(l) ----> Br-(aq)

Now to balance these two half reactions. In the first reaction, we must balance for both charge and matter. First, balance the Br atoms on both sides.

Br2(l) -----> 2BrO3-(aq)

Notice the missing oxygen on the reactant side. To compensate for this, we must add H2O to the reactant side:

Br2(l) + 6H2O(l) -----> 2BrO3-(aq)

Now, notice the products sides lacks H atoms, we must add H+ ions:

Br2(l) + 6H2O(l) -----> 2BrO3-(aq) + 12H+

Now, to balance for charge, we must add electrons (e-) to the proper side of the equation:

Br2(l) + 6H2O(l) ------> 2BrO3- + 12H+ + 10e-

Both sides of this half reaction are now balanced for atoms and charge (both sides have net 0 charge).

Now, let's do the same to the other half reaction:

Br2(l) ----> Br-(aq)

Balance Br atoms first

Br2(l) ---> 2Br-(aq)

Now balance for charge by adding electrons, e-:

Br2(l) + 2e- ---> 2Br-(aq)

Both sides have the same bromine atoms and both sides have the same charge of net -2.

This leaves us with out balanced half reactions:

Br2(l) + 2e- ---> 2Br-(aq)
Br2(l) + 6H2O(l) ------> 2BrO3- + 12H+ + 10e-

You can see that, in both cases, Br2(l) acts as both an oxidizing and reducing agent. In the end, Br- gains electrons while BrO3- releases electrons. Thus, Br- is the oxidizing agent (reduces) and BrO3- is the reducing agent (oxidizes). Br2(l) performs both of these functions.

Now, to add the half reactions together, we must remove the electrons from the half reactions. (Where have we ever seen electrons appear in overall reactions? Never, my friend, never!) To do so, we must multiply the first equation by 5, giving us 10e- on both sides:

5Br2(l) + 10e- ---> 10Br-(aq)
Br2(l) + 6H2O(l) ------> 2BrO3- + 12H+ + 10e-

Now add these two half reactions together and cancel out anything that appears on both sides of the equation (ex. electrons) and you will get:

5Br2(l) + Br2(l) + 6H2O(l) ----> 10Br-(aq) + 2BrO3- + 12H+
6Br2(l) + 6H2O(l) -----> 10Br-(aq) + 2BrO3- + 12H+

Thus, your net reaction is:
6Br2(l) + 6H2O(l) -----> 10Br-(aq) + 2BrO3- + 12H+

Notice how both sides are balanced for both atom numbers and charge? This is what we want in a redox reaction!

Now, this reaction we have is in ACIDIC solution. How do we know? Somewhere in our chemical redox reaction, we come across an H+... in fact, we come across 12! To make this acidic, we must neutralize the H+ by adding sufficient OH- ions to both sides of the reaction, thus forming H2O from the H+ ions. In our case, we'll need to add 12OH- to both sides, giving us:

6Br2(l) + 6H2O(l) + 12OH- -----> 10Br-(aq) + 2BrO3- + 12H+ + 12OH-
6Br2(l) + 6H2O(l) + 12OH- -----> 10Br-(aq) + 2BrO3- + 12H2O(l)
The H2O's on both sides can cancel, leaving us with
6Br2(l) + 12OH- -----> 10Br-(aq) + 2BrO3- + 6H2O(l)

You can once again check to see if both sides of the equation are balanced for atoms and for charge. You will find that, yes in fact, they are. This is our final redox reaction, balanced in basic solution:

6Br2(l) + 12OH- -----> 10Br-(aq) + 2BrO3- + 6H2O(l)

2)

Similarly

MnO4- is being reduced to Mn 2+ so the half equation is
MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

S2O3 2- is being oxidised to S4O6 2- & the half equation is
2S2O3 2- ---> S4O62- + 2e-

You balance the equations by multiplying the first half equation by 2 and the second by 5 to get the same number of e- on both sides (10e- on each), and overall it's

2MnO4- + 16H+ 10S2O3 2- ---> 2Mn2+ +8H2O + 5S4O62-

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