Electrochemistry
    7.0 Nature of Electrodes

7.0 Nature of Electrodes
There are various types of electrodes which are used in the chemical reactions, following are:

$(i)$ Gas Electrode: When an inert electrode is in the direct with the gas at a pressure of $1$ atm (unless specified) and other ions in the solution. For Example. Hydrogen gas electrode etc.
It is denoted as ${H^ + }/{H_2}(1atm)/Pt$ and the chemical reaction is $${H^ + } + {e^ - } \to \frac{1}{2}{H_2}_{(g)}$$
Inert electrode should be such that if the external voltage is changed by even very small amount from its equilibrium value, then reaction should proceed in either direction( forward or backward).Generally Platinum with fine coat of Platinum powder or Platinum black.



$(ii)$ Metal-insoluble salt Electrode: In this, metal electrode comes in contact with ions( generally anions). Common example is calomel electrode, which consists of mercury(Hg), mercurous chloride ($H{g_2}C{l_2}$) and saturated solution of potassium chloride ($KCl$). In both the cases, the concentrations of chloride ions are different. So the chemical reaction is
$$\frac{1}{2}H{g_2}C{l_2}_{(s)} + {e^ - } \to H{g_{(s)}} + C{l^ - }$$ or $$C{l^ - }/HgC{l_2}/Hg/Pt$$



$(iii)$ Metal-metal ion Electrode: Here metal electrode takes part in reaction. it is dipped into the solution containing its ion. Reactivity of the metal should be intermediate otherwise it will react with water and would not act as metal electrode. For Example: copper electrode is dipped into a solution of copper sulphate solution. Denoted as ${M^{n + }}/M$. And in case of copper, it is written as $C{u^{2 + }}(c)/Cu$ where $c$ is the concentration. The chemical reaction is $$C{u^{2 + }} + 2{e^ - } \to C{u_{(s)}}$$


$(iv)$ Redox Electrode (Oxidation-Reduction Electrode): Here inert electrode is dipped into a solution containing species which exist in two oxidation states. For Example. Platinum electrode is dipped into a solution of stannous and stannic, denoted by $$S{n^{4 + }},S{n^{2 + }}/Pt$$
and reaction is $S{n^{4 + }} + 2{e^ - } \to S{n^{2 + }}$. The solution can be neutral species also present in two oxidation states.
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