8.0 IUPAC Cell Representation and Convention

It is represented as Figure $$\underbrace {Zn/ZnS{O_4}}_{({C_1})}\parallel \underbrace {CuS{O_4}/Cu}_{({C_2})}$$

$(i)$ Tha anode (electrode) is written. After the anode, the electrolyte of the anode should be written with concentration. Here, concentration of ${ZnS{O_4}}$ is given as ${{C_1}}$ moles/litre.

$(ii)$ There is a slash $(/)$ put in between $Zn$ rod and the respective electrolyte which denotes the surface barrier between these two as they exist in different phases.

$(iii)$ Then after two vertical lines are drawn which are used to represent the salt bridge

$(iv)$ Now we need to write the electrolyte of right side compartment or cathode half cell which is ${CuS{O_4}}$ in this case. And its concentration is also written (here it is ${{C_2}}$ moles/litre).

$(v)$ Now again a single slash(/) is drawn to show the barrier between electrode and electrolyte.

$(vi)$ And finally we write the cathode electrode of the cathode half cell.

So we can say that oxidation half reaction are always written on left- side and cathode or reduction half reaction are written on right side.

In case of gas electrode, gas is to be indicated after the electrode in case of anode and before the electrode in case of cathode. $$Pt,C{l_2}/C{l^ - }\ \ \ or\ \ \ C{l^ - }/C{l_2},Pt$$

$(vii)$ The cell terminals are at the extreme ends in this cell notation. Half reaction involves a gas, and an inert material such as Platinum serves as a terminal and also as electrode on which reaction occurs. Platinum catalyses the reaction but does not involve in the reaction. Hydrogen bubbles on a platinum plate which is immersed in acidic medium.

So the cathodic half reaction is $$2{H^ + }_{(aq)} \to {H_{2(g)}}$$

Representation of hydrogen electrode, acts as cathode $${H^ + }_{(aq)}/{H_{2(g)}}/Pt$$ or $${H^ + }_{(aq)}/{H_{2(g)}},Pt$$ or $${H^ + }_{(aq)}/Pt, {H_2}$$

If this electrode acts as anode, then reverse the notation is $$Pt/{H_{2(g)}}/{H^ + }_{(g)}$$

Note: The half cell which has higher reduction potential acts as cathode, and the electrode with lower reduction potential acts as anode.

Question 3. Write the cell reaction for the voltaic cell $$T{l_{(s)}}/T{l_{(aq)}}//S{n^{2 + }}_{(aq)}/S{n_{(s)}}$$

Solution: The half cell reactions are $$\begin{equation} \begin{aligned} T{l_{(s)}} \to T{l^ + }_{(aq)} + {e^ - }...(i) \\ S{n^{2 + }}_{(aq)} + 2{e^ - } \to S{n_{(s)}}...(ii) \\\end{aligned} \end{equation} $$

Now we need to decide which is cathode , which is anode.As we know that on anode, oxidation occurs so reaction $(i)$ is anode and other is cathode .

Multiplying $(i)$ by $2$ and then adding with $(ii)$, we get $$2T{l_{(s)}} + S{n^{2 + }}_{(aq)} \to 2T{l^ + }_{(aq)} + S{n_{(s)}}$$

To properly or fully specify the voltaic cell, we need to give the concentration of species or ions and also pressure of the gas if gaseous electrode is used. These things we will study in further topics.