Chapter 8.1 - Electrochemical Series

In the previous section we saw that, redox reactions can be explained on the basis of electron transfer. So such reactions will depend on the 'readiness' of one of the reactants to donate electrons. We know that, elements with metallic character, tends to donate electrons. But what if more than one metals are present in a reaction? Which metal will donate electrons? In this section, we try to find answers to those questions.

Let us see an experiment in which two metals, copper and zinc are involved. It can be written in 7 steps:
1. Take an aqueous solution of copper nitrate in a beaker.
2. Place a strip of metallic zinc as shown in fig.8.1(a) below:

Fig.8.1

3. Wait for one hour. We will see that:
    ♦ The strip becomes coated with a reddish substance.
         ✰ This is shown in fig.b
    ♦ The blue color of the solution disappears.
         ✰ This is shown in fig.c
4. The reason can be written in four steps:
(i) The blue color of the solution was due to the presence of Cu²⁺ ions.
• The Cu²⁺ ions are formed by the dissociation of copper nitrate molecules in aqueous solution.
(ii) When the zinc strip is place in the solution, each Cu²⁺ ion receives two electrons from the Zn atoms.
• The Cu²⁺ thus becomes Cu, which is pure copper.
(iii) This pure copper gets deposited on the surface of the zinc strip. This is the reason for the reddish coating.
(iv) As time passes, number of Cu²⁺ ions decreases. Thus the blue color of the solution becomes pale blue and finally, it becomes color less.
5. Though the solution is now color less, Zn²⁺ ions are present in it. This can be confirmed by passing hydrogen sulfide gas through the color less solution. We will get a white precipitate of zinc sulfide.
(Some of the remaining Cu²⁺ ions may also react with the hydrogen sulfide to form copper sulfide. This copper sulfide is dark black in color. So the precipitate will be a mixture of zinc sulfide and copper sulfide. This will make the precipitate look grayish. In order to see the pure white of zinc sulfide, we add a few drops of ammonia to the resulting solution. This will dissolve the copper sulfide. Then we will be able to see the pure white zinc sulfide)
6. We can demonstrate the electron transfer through half reactions:
    ♦ Zn (s) → Zn²⁺ (aq) + 2e^-
    ♦ Cu²⁺ (aq) + 2e^- → Cu (s)
• Adding the two equations, we get:
Zn (s) + Cu²⁺ + 2e^- → Zn²⁺ + 2e^- + Cu (s)
• The electrons on both sides will cancel each other. Thus we get:
Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)
7. Based on the above steps, we can write a summary:
• Both copper and zinc are metals. They both tend to donate electrons.
• But in the present case, copper was forced to accept electrons.
• This is because, zinc is a better electron donor than copper.

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◼ We saw that:
In case of a competition between zinc and copper (to find who is a better electron donor), zinc will win.
• We can confirm this by another experiment. It can be written in 6 steps:
1. Take an aqueous solution of zinc sulfate in a beaker.
2. Place a strip of metallic copper.
3. The zinc sulfate solution will contain Zn²⁺ ions. We would expect those ions to accept electrons from the metallic copper atoms. Thus we would expect metallic zinc to get deposited on the strip.
• Also we would expect the presence of Cu²⁺ ions in the solution.
4. But neither will happen:
    ♦ No zinc will get deposited.
    ♦ Not even a single Cu²⁺ ion will become available in the solution.
5. The absence of Cu²⁺ can be confirmed by passing hydrogen sulfide gas.
• If there is any Cu²⁺ ions, black cupric sulfide will form.
• Cupric sulfide is very insoluble. So even it’s small presence can be easily detected. Yet we find no trace of it. That means, no Cu²⁺ is formed.
6. Thus we can confirm that, Cu cannot become Cu²⁺ in the presence of Zn
• So we can write the order in which the two metals are able to donate electrons: Zn > Cu
◼ That is., zinc has a greater capacity to donate electrons when compared to copper

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Next we want to know the metal which comes after Cu. For that, we perform another experiment. It can be written in 8 steps:
1. Take an aqueous solution of silver nitrate in a beaker.
2. Place a strip of metallic copper as shown in fig.8.2(a) below:

Fig.8.2

3. Wait for one hour. We will see that:
    ♦ The strip becomes coated with a silvery substance.
         ✰ This is shown in fig.b
    ♦ The solution gradually becomes blue in color.
         ✰ This is shown in fig.c
4. The reason can be written in four steps:
(i) The solution initially contains Ag⁺ ions.
• The Ag⁺ ions are formed by the dissociation of silver nitrate molecules in aqueous solution.
(ii) When the copper strip is place in the solution, two Ag⁺ ions receive one electron each from a Cu atom.
• The Ag⁺ thus becomes Ag, which is pure silver.
(iii) This pure silver gets deposited on the surface of the copper strip. This is the reason for the silvery coating.
(iv) After donating electrons, the Cu atoms go into the solution as Cu²⁺. As time passes, number of Cu²⁺ ions in the solution increases. Thus the blue color of the solution increases.
5.We can demonstrate the electron transfer through half reactions:
    ♦ Cu (s) → Cu²⁺ (aq) + 2e^-
    ♦ 2Ag⁺ (aq) + 2e^- → 2Ag (s)
• Adding the two equations, we get:
Cu (s) + 2Ag⁺ (aq) + 2e^- → Cu²⁺ (aq) + 2e^- + 2Ag (s)
• The electrons on both sides will cancel each other. Thus we get:
Cu (s) + 2Ag⁺ (aq) → Cu²⁺ (aq) + 2Ag (s)
6. Based on the above steps, we can write a summary:
• Both silver and copper are metals. They both tend to donate electrons.
• But in the present case, silver was forced to accept electrons.
• This is because, copper is a better electron donor than silver.
7. So now we know that, silver comes behind copper in the list. We get:
Zn > Cu < Ag
8. We can write:
   ♦ Zn has the ability to reduce Cu²⁺ to Cu
   ♦ Cu has the ability to reduce Ag⁺ to Ag

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• Our list contains only three metals. But scientists have prepared a detailed list in which many more metals are present.
• It is known as the metal activity series or the electrochemical series.
• Such a list becomes useful while designing Galvanic cells. The Galvanic cells depend on transfer of electrons between metals to produce electric current. We will study them in detail in class 12.

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◼ We saw two cases:
    ♦ Zinc has greater capacity to donate electrons than copper.
    ♦ Copper has greater capacity to donate electrons than silver.
• There are some metals which have equal capacities.
    ♦ If we use such metals in the Galvanic cells, current will not flow.
Let us see an example of such metals. It can be written in 7 steps:
1. Take an aqueous solution of nickel sulfate in a beaker.
2. Place a strip of metallic cobalt.
3. As usual,
    ♦ The Co will donate electrons and go into the solution as Co²⁺.
    ♦ The Ni²⁺ ions in the solution will accept those electrons and become Ni.
4. We can demonstrate the electron transfer through half reactions:
    ♦ Co (s) → Co²⁺ (aq) + 2e^-
    ♦ Ni²⁺ (aq) + 2e^- → Ni (s)
• Adding the two equations, we get:
Co (s) + Ni²⁺ (aq) + 2e^- → Co²⁺ (aq) + 2e^- + Ni (s)
• The electrons on both sides will cancel each other. Thus we get:
Co (s) + Ni²⁺ (aq) → Co²⁺ (aq) + Ni (s)
5. We would expect all the Ni²⁺ ions in the solution to become Ni atoms. But this does not happen.
• Experiments show that:
At equilibrium, both Ni²⁺ ions and Co²⁺ ions are present in moderate concentrations.
6. That means, cobalt do not have a greater capacity then nickel, so as to force nickel to accept electrons.
7. Neither does nickel have a greater capacity than cobalt, so as to force cobalt to accept electrons. This becomes evident when we consider the reverse reaction in (4).

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• We have seen the basic details about electrochemical series. In the next section, we will see oxidation number.


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