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Metal–metal ion half cell

In this type of half-cell, the metal rod is dipped in a solution of the corresponding metal ion having concentration c M.
 
Description: 31082.png
Fig. 10.1
 
The half cell shown in Fig. 10.1 when functions as anode can be represented as
 
M(s) | Mn+ (c M)
 
The anodic half cell reaction is represented as
 
M(s) Mn+(aq) + ne
 
The representation of the half cell when it functions as cathode is Mn+ (c M) | M(s).
 
The cathodic half cell reaction is Mn+(aq) + ne M(s).
 
Examples of this type of half cell when used as cathode are Cu2+ | Cu, Zn2+ | Zn, Ag+ | Ag, Sn2+ | Sn, etc. The half cells or electrodes used in Daniel cell are of the metal–metal ion type.

Gas–gas ion half cell

In a gasgas ion half cell, a gas is bubbled into a solution of the gas ion, with a platinum rod having a sheet coated with platinum black being dipped in the gas ion solution. Platinum is used for making electrical contact because the gases are non-conducting and the platinum does not react with gas ions as it is an unreactive metal. Platinum black is used over the platinum sheet, which gives it a large surface area for the adsorption of gases. The half cell shown in Fig. 10.2 when functions as anode can be represented as
 
Pt | Gas (P atm) | Gas ion (c M).
Description: 31093.png
Fig. 10.2
 
Let us consider a hydrogen gas half cell, functioning as anode. It is represented as
 
Pt | H2 (P atm) | H+ (c M)
 
and the anodic half-cell reaction is
 
H2(Pt) 2H+(aq) + 2e(Pt)
 
The oxygen gas half cell when functions as anode can be represented as
 
Pt | O2 (P atm) | OH (c M)
 
and the anodic half cell reaction is
 
2OH ½O2 + H2O + 2e
 
The gas–gas ion half cell when functions as cathode can be represented as
 
Gas ion (c M) | Gas (P atm) | Pt
 
Let us consider a hydrogen half cell, functioning as cathode. It is represented as
 
H+ (c M) | H2 (P atm) | Pt
 
and the cathodic half cell reaction is
 
2H+(aq) + 2e (Pt) H2 (Pt)
 
The hydrogen electrode is called standard hydrogen electrode when the concentration of H+ ion is 1 M and the pressure of H2 gas is 1 atm.

Redox half cell

In redox half cell, a platinum rod is dipped in a solution containing two different oxidation states of a metal ion Description: 31101.png where n2 is greater than n1).
 
The half cell shown in Fig. 10.3 when functions as an anode can be represented as
 
Description: 34040.png
 
and the anodic half cell reaction is
 
Description: 34031.png
 
The redox half cell when functions as a cathode can be represented as
 
Description: 34024.png
 
and the catholic half cell reaction is
 
Description: 34017.png
 
Description: 31147.png
Fig. 10.3
 
Examples of this type of half cell when used as anode are Pt | Cr2+ (c1 M), Cr3+ (c2 M); Pt | Fe2+ (c1 M), Fe3+ (c2 M); Pt | Sn2+ (c1 M), Sn4+ (c2 M), etc. In such half cells, the oxidizing and reducing agents are both metal ions.

Metal–insoluble metal salt–anion half cell

Let us take an example of such a half cell, which functions as cathode, as Cl (c M) | AgCl | Ag. The assembly for this half cell is made by dipping a rod of silver coated with a paste of saturated AgCl at the bottom in a solution of ionic electrolyte such as KCl, NaCl, or HCl. The cation part in the electrolyte must have standard reduction potential less than the standard reduction potential of the cation of the metal salt (see Fig. 10.4).
 
Description: 31158.png
Fig. 10.4
 
The cathodic half cell reaction is
 

 
Applying Nernst equation to the net half cell reaction, we get
 
(AgCl and Ag do not appear in the expression as they are pure solids.)
 
Now, if we are given Description: 31245.png concentration of Cl, and KSP of AgCl, how can we calculate the half cell potential Description: 31253.png for the reaction AgCl(s) + e Ag(s) + Cl.
 
In Nernst equation, can Description: 31262.png be replaced for Description: 31270.png? Let us see.
 
 
It is evident that Description: 31309.png is not equal to Description: 31318.png as Description: 31327.png is not zero. Thus, Description: 31334.png is not equal to Description: 31342.png and we cannot replace Description: 31351.png by Description: 31360.png
 
But using Hess’s law, we can calculate Description: 31368.png as
 
Description: 31381.png Description: 31389.png
 
Dividing this equation by –F, we get
 
Description: 31426.png
 
Substituting the value of Description: 31434.png in Eq. (i), we get
Description: 31443.png
Description: 31452.png
 
Since, Description: 31461.png concentration of Cl and KSP of AgCl are given, the reduction potential Description: 31469.png for the reaction, AgCl(s) + e Ag(s) + Cl can be calculated.




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