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How do you calculate electrochemical cell potential?

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Hint: There two parts in the electrochemical reaction that is oxidation half-reaction and reduction half-reaction. There are two factors for calculating the electrochemical cell potential i.e., standard emf of the cathode and the standard emf of the anode.


Complete Solution :
- The electrochemical cell potential of the reaction is also known as the standard emf of the cell. An electrochemical cell is based on a reaction which can be split into two half reactions i.e. oxidation half-reaction and reduction half-reaction.
- Oxidation half-reaction is the reaction in which the electrons are lost and the cation is formed.
-Reduction half-reaction is the reaction in which the electrons are gained and the metal is formed.
Let us take an example of the reaction of zinc and silver is given below:
$Zn+2A{{g}^{+}}\to Z{{n}^{2+}}+2Ag$
So, in this reaction give above, the oxidation half-reaction will be of zinc and the reaction will be:
$Zn\to Z{{n}^{2+}}+2{{e}^{-}}$
The reduction half-reaction will be of silver and the reaction will be:
$2A{{g}^{+}}+2{{e}^{-}}\to 2Ag$
The cell of this reaction will be represented as:
$Zn|Z{{n}^{2+}}||A{{g}^{+}}|Ag$

The right side of the cell will be the cathode and the left side of the cell will be the anode.
The standard emf of the cell of the electrochemical reaction is equal to the difference between the standard emf of the cathode and the standard emf of the anode. The formula is written as:
$E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$

Note: If the reaction is not in the normal condition then the formula can be used is:
${{E}_{cell}}=E_{cell}^{\circ }-\dfrac{RT}{nF}\ln Q$
- Where R is the gas constant, T is the temperature, n is the number of electrons in the reaction, F is Faraday's number, and Q is the reaction quotient.