 # what is the procedure of cell reaction??????? 11 years ago

Dear pallavi,

Standard electrode potentials are usually tabulated as reduction potentials. However, the reactions are reversible and the role of a particular electrode in a cell depends on the relative oxidation/reduction potential of both electrodes. The oxidation potential for a particular electrode is just the negative of the reduction potential. A standard cell potential can be determined by looking up the standard electrode potentials for both electrodes (sometimes called half cell potentials). The one that is smaller will be the anode and will undergo oxidation. The cell potential is then calculated as the sum of the reduction potential for the cathode and the oxidation potential for the anode. $\mbox{E}^{o}_{cell}=\mbox{E}^{o}_{red}(cathode)-\mbox{E}^{o}_{red}(anode) = \mbox{E}^{o}_{red}(cathode)+\mbox{E}^{o}_{oxi}(anode)$

For example, the standard electrode potential for a copper electrode is: $\mbox{Cell diagram}\,$ $\mbox{Pt}(s)|\mbox{H}_{2}(1 atm)|\mbox{H}^{+}(1 M)||\mbox{Cu}^{2+}(1 M)|\mbox{Cu}(s)\,$ $\mbox{E}^{o}_{cell}=\mbox{E}^{o}_{red}(cathode)-\mbox{E}^{o}_{red}(anode)$

At standard temperature, pressure and concentration conditions, the cell's EMF  is 0.34 V. by definition, the electrode potential for the SHE is zero. Thus, the Cu is the cathode and the SHE is the anode giving $\mbox{E}_{cell}=\mbox{E}^{o}_{\mbox{Cu}^{2+}/\mbox{Cu}}-\mbox{E}^{o}_{\mbox{H}^{+}/\mbox{H}_{2}}$

Or, $\mbox{E}^{o}_{\mbox{Cu}^{2+}/\mbox{Cu}} = \mbox{0.34 V}$

So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Please feel free to post as many doubts on our discussion forum as you can.we will get you the answer and detailed  solution very  quickly.

All the best.

Regards,