Electrochemistry
9.0 Standard Cell EMF and Standard Reduction Potential
9.0 Standard Cell EMF and Standard Reduction Potential
A cell emf is a measure of driving force of the reaction. The reaction occurs in two half cells referred as half oxidation reaction and half reduction reaction.
If we construct a table of reduction potential, we will get a list of strengths of oxidizing agents and then also a way to calculate cell emf.
Oxidation half reaction is reverse of reduction half reaction of same species. So oxidation potential for oxidation half reaction is negative of reduction potential of reduction half reaction.
Oxidation potential for oxidation half reaction $=\ -$ reduction potential of reduction half reaction
So we can say that either we should formulate a table of oxidation potential or reduction potential. And accordingly we can find out the other.
Electrode potential is denoted as ${E_{cell}}$
Consider an example of $Zn-Cu$
$$Z{n_{(s)}}/Z{n^{2 + }}_{(aq)}//C{u^{2 + }}_{(aq)}/C{u_{(s)}}$$
So writing the oxidation and reductionreaction separately as two half cell reaction:
$$\begin{equation} \begin{aligned} Z{n_{(s)}} \to Z{n^{2 + }}_{(aq)} + 2{e^ - }...(i) \\ C{u^{2 + }}_{(aq)} + 2{e^ - } \to C{u_{(s)}}...(ii) \\\end{aligned} \end{equation} $$
Reaction $(i)$ is an oxidation reaction so oxidation potential for this reaction be ${E_{oxd}}$ or ${E_{Zn}}$ and reaction $(ii)$ is a reduction reaction so reduction potential for this reaction be ${E_{red}}$ or ${E_{Cu}}$.
So emf is the sum of oxidation potential and reduction potential and is written as
$${E_{cell}}={E_{oxd}}+{E_{red}}$$$${E_{cell}}={E_{cathode}}-{E_{anode}}$$$${E_{cell}}={E_{right}}-{E_{left}}$$
The electrode potential is an intensive property.This means that value of potential is independent of amount of species in the reaction. So electrode potential for the reaction $$Z{n_{(s)}} \to Z{n^{2 + }}_{(aq)} + 2{e^ - }$ and $Z{n_{(s)}} \to Z{n^{2 + }}_{(aq)} + 2{e^ - }$$
are same.
Note: If ${E_{cell}}$ is positive then the reaction is spontaneous.
If in any cell reaction, more than $1$ electron are involved then standard electrode potential of the cell remains same but standard free energy will get change depending on the value of $"n"$ where n is the number of electrons involved in the reaction.
Standard Electrode Potential
The emf of the cell depends on the concentrations of the species and the temperature of the cell.For making a electro-chemical series or table, we should choose some standard thermodynamic conditions. The standard emf, ${E^0}_{oxd}$ is the emf of the cell which is operating under standard state conditions (solute concentrations are each $1\ M$, gas pressures are each $1\ atm$ and temperature is generally ${25^ \circ }$.
Note: Wherever subscript $0$ is used, it means variables are for standard conditions.
If we tabulate the electrode potential then we can calculate emf from them. So from a small table of about $40$ electrode potential, it can give emf of nearly $800$ voltaic cells.
It is not possible to measure the emf of the single electrode, only emfs can be found out . When we need to calculate the emf of the cell constructed from various electrodes which are in turn connected to one particular electrode which you choose as a reference electrode. We arbitrary assign the sign or the value of this reference electrode, a potential of zero and then get the potential for the electrode which are connected to form a cell. By convention, reference electrode chosen is standard hydrogen electrode.
Let us understand how these values are found of standard electrode potential. let us find out the standard electrode potential for the zinc electrode which is connected to a standard hydrogen electrode. When we measure the emf of the cell by voltmeter and we obtain $0.76$V, where zinc electrode acts as a anode.
The cell is $$Z{n_{(s)}}/Z{n^{2 + }}_{(aq)}//{H^ + }_{(aq)}/{H_{2(g)}}/Pt$$
The half reactions are $$Z{n_{(s)}} \to Z{n^{2 + }}_{(aq)} + 2{e^ - }$$
The cell emf is the sum of the half cell potentials.
Put $0.76V$ for the cell emf and $0.00V$ for the standard hydrogen potential, which gives ${E_{Cu}}$ =$-0.76V$.
Moving in this way we can obtain electrode potential for a series of half -cell reactions. Table shown below lists the standard electrode potential for a half cell series at ${25^ \circ }C$.
These metals are more reactive than hydrogen | Potassium | $K$ | Most Reactive metal |
| Sodium | $Na$ | ||
| Calcium | $Ca$ | ||
| Magnesium | $Mg$ | ||
| Aluminium | $Al$ | ||
| Zinc | $Zn$ | ||
| Iron | $Fe$ | ||
| Tin | $Sn$ | ||
| Lead | $Pb$ | ||
| [Hydrogen] | $[H]$ | ||
These metals are less reactive than hydrogen | Copper | $Cu$ | |
| Mercury | $Hg$ | ||
| Silver | $Ag$ | ||
| Gold | $Au$ | Least Reactive metal |
Electro-chemical Series table
Negative values of electrode reduction potential signifies that an electrode when joins with SHE acts as anode and oxidation occurs on this electrode. In same way, we can say that positive sign of electrode reduction potential signifies that an electrode when joins with SHE acts as cathode and reduction occurs on this electrode.
