Assertion (A) Copper sulphate can be stored in zinc vessel.
Reason (R) Zinc is less reactive than copper.
Consider the figure and answer the following questions.
(i) Cell ' $A$ ' has $\mathrm{E}_{\text {cell }}=2 \mathrm{~V}$ and Cell ' B ' has $\mathrm{E}_{\text {cell }}=1.1 \mathrm{~V}$ which of the two cells 'A' or 'B' will act as an electrolytic cell. Which electrode reactions will occur in this cell?
(ii) If cell ' $A$ ' has $\mathrm{E}_{\text {cell }}=0.5 \mathrm{~V}$ and cell ' $B^{\prime}$ has $\mathrm{E}_{\text {cell }}=1.1 \mathrm{~V}$ then what will be the reactions at anode and cathode?
(i) Cell ' $B$ ' will act as electrolytic cell due to its lesser value of emf.
The electrode reactions will be
At cathode $$\quad\mathrm{Zn}^{2+}+2 e^{-} \longrightarrow \mathrm{Zn}$$
At anode $$\quad \mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2 e^{-}$$
(ii) If cell ' $B$ ' has higher emf, it acts as galvanic cell.
Now it will push electrons into cell ' $A$ '
In this case, the reactions will be
$$\begin{aligned} \mathrm{Zn} & \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \text {(At anode) } \\ \mathrm{Cu}^{2+}+2 e^{-} & \longrightarrow \mathrm{Cu} \text { (At cathode) } \end{aligned}$$
Consider figure from the above question and answer the questions (i) to (vi) given below.
(i) Redraw the diagram to show the direction of electron flow.
(ii) Is silver plate the anode or cathode?
(iii) What will happen if salt bridge is removed?
(iv) When will the cell stop functioning?
(v) How will concentration of $\mathrm{Zn}^{2+}$ ions and $\mathrm{Ag}^{+}$ions be affected when the cell functions?
(vi) How will the concentration of $\mathrm{Zn}^{2+}$ ions and $\mathrm{Ag}^{+}$ ions be affected after the cell becomes 'dead'?
(i) Electrons move from Zn to Ag as $E^{\circ}$ is more negative for Zn , so Zn undergoes oxidation and $\mathrm{Ag}^{+}$undergoes reduction.
(ii) Ag is the cathode as it is the site of reduction where $\mathrm{Ag}^{+}$takes electrons from medium and deposit at cathode.
(iii) Cell will stop functioning because cell potential drops to zero. At $E=0$ reaction reaches equilibrium.
(iv) When $E_{\text {cell }}=0$ because at this condition reaction reaches to equilibrium.
(v) Concentration of $\mathrm{Zn}^{2+}$ ions will increase and concentration of $\mathrm{Ag}^{+}$ions will decrease because Zn is converted into $\mathrm{Zn}^{2+}$ and $\mathrm{Ag}^{+}$is converted into Ag .
(vi) When $E_{\text {cell }}=0$ equilibrium is reached and concentration of $\mathrm{Zn}^{2+}$ ions and $\mathrm{Ag}^{+}$will not change.
What is the relationship between Gibbs free energy of the cell reaction in a galvanic cell and the emf of the cell? When will the maximum work be obtained from a galvanic cell?
If concentration of all reacting species is unity, then $E_{\text {cell }}=E_{\text {cell }}^{\circ}$ and $\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}$ where, $\Delta_r G^{\circ}$ is standard Gibbs energy of the reaction
$$\begin{aligned} E_{\text {cell }}^{\circ} & =\text { emf of the cell } \\ n F & =\text { charge passed } \end{aligned}$$
If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly.
The reversibly work done by a galvanic cell is equal to decrease in its Gibbs energy.
$$\Delta_r G=-n F E_{\text {cell }}$$
As $E_{\text {cell }}$ is an intensive parameter but $\Delta_r G$ is an extensive thermodynamic property and the value depends on $n$.
For reaction, $\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(\mathrm{s})$ in a galvanic cell.
$$\Delta_r G=-2 F E_{\text {cell }} \quad[\text { Here, } n=2]$$