Question
What is cell potential or emf of a cell ?
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Answer
Cell potential or emf of a cell : It is defined as the potential difference between two electrodes, responsible for an external flow of electrons from the left hand electrode at higher potential (anode), to the right hand electrode at lower potential (cathode), when connected to form an electrochemical or galvanic cell.
Since there is oxidation reaction at left hand electrode (LHE) or anode and reduction reaction at right hand electrode (RHE) or cathode, emf of the galvanic, Ecell, is given by
$E _{\text {cell }}=\left( E _{\text {oxi }}\right)_{\text {anode }}+\left( E _{\text {red }}\right)_{\text {cathode }}$
Since by IUPAC conventions, generally reduction potentials are used, hence, for the given cell,
$\left(\because E_{\text {oxi }}=-E_{\text {red }}\right)$
$\therefore E _{\text {cell }}=\left( E _{\text {red }}\right)_{\text {cathode }}-\left( E _{\text {red }}\right)_{\text {anode }}$
Similarly, standard emf of the cell, $E^0$ cell is given by
$E ^0{ }_{\text {cell }}=\left( E _{\text {red }}^0\right)_{\text {cathode }}-\left( E _{\text {red }}^0\right) \text { anode }$
Since there is oxidation reaction at left hand electrode (LHE) or anode and reduction reaction at right hand electrode (RHE) or cathode, emf of the galvanic, Ecell, is given by
$E _{\text {cell }}=\left( E _{\text {oxi }}\right)_{\text {anode }}+\left( E _{\text {red }}\right)_{\text {cathode }}$
Since by IUPAC conventions, generally reduction potentials are used, hence, for the given cell,
$\left(\because E_{\text {oxi }}=-E_{\text {red }}\right)$
$\therefore E _{\text {cell }}=\left( E _{\text {red }}\right)_{\text {cathode }}-\left( E _{\text {red }}\right)_{\text {anode }}$
Similarly, standard emf of the cell, $E^0$ cell is given by
$E ^0{ }_{\text {cell }}=\left( E _{\text {red }}^0\right)_{\text {cathode }}-\left( E _{\text {red }}^0\right) \text { anode }$
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