Help in chemistry pleeease

Question

97)

Consider the following unbalanced redox reaction.

MnO4^- (aq) Zn (s) ----> Mn^2 (aq) Zn^2 (aq)

Balance the equation and determine how much of a 0.500 M KMn)4 solution is required to completely dissolve 2.85 g of Zn.

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Please explain so I can understand in the future... thanks :)

Answer 1

First you need to balance the half reactions:

MnO4^-    ----->   Mn^2+

The oxidation state of Manganese decreases from 7+  to 2+. In order for that to have occured, MnO4^- had to reduce by gaining 5 electrons.

MnO4^-  + 5e^-    ----->   Mn^2+  

Balance oxygens with water:

MnO4^-  + 5e^-    ----->   Mn^2+   + 4H20

Balance hydrogens with acid (H+):

8H^+  +   MnO4^-  + 5e^-    ----->   Mn^2+   + 4H20

This half reaction is balance, now for the second one.

Before we get into the second one, since the Manganese in MnO4^- was reduced, we know zinc must be oxidized otherwise no redox reaction would be occuring in the first place.

Zn  ---->   Zn^2+

This reaction is a lot easier to balance... only the charge needs to be balanced by electrons and you can see Zn is in fact oxidizing.

Zn  ---->   Zn^2+   + 2e^-

Now we need to multiply BOTH half reaction so the number of electrons is equal for both. The Zn half reaction has 2e- and the Mn half reaction has 5. The lowest multiple of both 2 and 5 is 10 so we must multiply the Mn half reaction by 2 and the Zn half reaction 5.

5 * [Zn  ---->   Zn^2+   + 2e^- ]

2 * [8H^+  +   MnO4^-  + 5e^-    ----->   Mn^2+   + 4H20]

This becomes:

5Zn  ---->   5Zn^2+   + 10e^-

16H^+  +   2MnO4^-  + 10e^-    ----->   2Mn^2+   + 8H20

Now we can combine them and you see the electrons cancel from both sides. They MUST because you cannot have electrons spontaneously leaving a reducing atom without another oxidizing atom to accept it.

The balance redox is:

5Zn +  16H^+  +   2MnO4^-    ----->   2Mn^2+   + 8H20 +   5Zn^2+  

And if you need a little bit more information about what is going on, Zn is oxidizing and acting as your reducing agent (or reductant) and MnO4^- reduces with Mn reducing and acting as the oxidizing agent (or oxidant).

Alright the second part of the question. This question is really a redox titration and is very similar to an acid base titration in terms of what happens. In this instance, we assume Zinc is the analyte and we are titrating Zinc with KMnO4^- solution. In case you were wondering, we will know when this titration will end because the solution will turn purple. KMnO4^- is purple in solution, the second it is dripped into Zn, the MnO4^- ion is reduced and turned to Mn^2+ (which is colorless). When all the Zn is used up, then KMnO4 will stop reducing and solution will turn purple.

If we have 2.85 g of Zn, then we have 2.85 g * 1n/(65.38g) = 0.0436 moles of Zn.

Looking at the balanced redox reaction, for every 5 moles of zinc, 2 moles of MnO4^-. This means for every 0.436 moles of Zn, (0.0436) * (2 moles MnO4^-)/(5 moles Zn) = 0.0174 moles of MnO4^- are used.

So that means 0.0174 moles of MnO4^- are needed for this reaction.

We are told the concentration of KMnO4^- is 0.500 M (the concentration of MnO4^- is the same concentration because KMnO4^- dissociates in K+ and MnO4^- in solution)

Concentration * Volume = moles

0.500 M MnO4^- * Volume = 0.0174 moles MnO4^-

Volume = 0.0348 Liter

You need 0.0348 liters (or 34.8 mL) of KMnO4