Question

Calculate the standard potential of the zinc–mercuric oxide cell shown in Figure 17.18. (Hint: The easiest way to proceed is to calculate$\Delta G_{}^0$for the corresponding overall reaction, and then find DE° from it.) Take$\Delta G_f^0$$\left[Zn{\left(OH\right)}_2\left(s\right)\right]=-\mathbf{553}.\mathbf{5}\mathbf{kJ}{\mathbf{mol}}^{-1}$

Short answer

The standard potential of zinc-mercury oxide (ZnO) cells is 1.34 V.

Step-by-step solution

Zinc-mercury oxide reaction.

Zinc-mercury oxide cell is a primary cell that cannot be used again once discharged. A mixture of mercuryand zinc (Zn) acts as anode and zinc (Zn) oxide from Zn to${\text{ZN}}^{\text{2+}}$at anode.

Steel rods in contact with solidmercury (II) oxide (HgO)act as a cathode,and mercury is reduced from${\text{Hg}}^{\text{2+}}$inmercury oxide (HgO) to mercury at the cathode. 45% potassium hydroxide (KOH) is taken as the electrolyte.

The half-cell reaction at the anode is shown below

$\text{Zn}\left(\text{s}\right){\text{+2OH}}^{\text{-}}\to \text{Zn}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right){\text{+2e}}^{\text{-}}$

The half-cell reaction of the cathode is shown below,

$\text{HgO}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right){\text{+2e}}^{\text{-}}\to \text{Hg}\left(\text{l}\right){\text{+2OH}}^{\text{-}}$

Here, the overall reaction of the cell is shown below,

$\text{Zn}\left(\text{s}\right)\text{+HgO}\left(\text{s}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\to \text{Zn}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)\text{+Hg}\left(\text{l}\right)$--- (a)

The standard free energy change.

The standard cell potential$\left({\text{ΔE}}_{\text{cell}}^{\text{o}}\right)$for equation (a) is calculated using the relation between the standard free energy change$\left({\text{ΔG}}^{\text{o}}\right)$, as follows;

${\text{ΔG}}^{\text{o}}{\text{=-nFΔE}}_{\text{cell}}^{\text{o}}$--- (b)

Here,

n is the number of moles of the electron.

F is the faraday constant.

Theis $\left({\text{ΔG}}^{\text{o}}\right)$calculated by using the below formula,

${\text{ΔG}}^{\text{o}}{\text{=ΔG}}_{\text{f.product}}^{\text{o}}{\text{-ΔG}}_{\text{f.reactant}}^{\text{o}}$ --- (c)

Here,

${\text{ΔG}}_{\text{f.product}}^{\text{o}}$is the standard free energy of formation for the product.

${\text{ΔG}}_{\text{f.reactant}}^{\text{o}}$is the standard free energy of formation for reactants.

Now, modify equation (c) for equation (a) we get,

${\text{ΔG}}^{\text{o}}\text{=}\left({\text{ΔG}}_{\text{f.Zn}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)}^{\text{o}}{\text{+ΔG}}_{\text{f.Hg}\left(\text{l}\right)}^{\text{o}}\right)\text{-}\left({\text{ΔG}}_{\text{f.Zn}\left(\text{s}\right)}^{\text{o}}{\text{+ΔG}}_{\text{f.HgO}\left(\text{s}\right)}^{\text{o}}{\text{+ΔG}}_{{\text{f.H}}_{\text{2}}\text{O}\left(\text{l}\right)}^{\text{o}}\right)$--- (d)

Here, the subscripts denote the reactants and products.

Now,put${\text{ΔG}}_{\text{f.Zn}{\left(\text{OH}\right)}_{\text{2}}\left(\text{s}\right)}^{\text{o}}{\text{=-553.5kJmol}}^{\text{-1}}$

$\begin{array}{l}{\text{ΔG}}_{\text{f.Hg}\left(\text{l}\right)}^{\text{o}}{\text{=0kJmol}}^{\text{-1}}\\ {\text{ΔG}}_{\text{f.Zn}\left(\text{s}\right)}^{\text{o}}{\text{=0kJmol}}^{\text{-1}}\\ {\text{ΔG}}_{\text{f.HgO}\left(\text{s}\right)}^{\text{o}}{\text{=-58.56kJmol}}^{\text{-1}}\\ {\text{ΔG}}_{{\text{f.H}}_{\text{2}}\text{O}\left(\text{l}\right)}^{\text{o}}{\text{=-237.18kJmol}}^{\text{-1}}\end{array}$

Putting the values in equation (d);

$\begin{array}{c}{\text{ΔG}}^{\text{o}}\text{=}\left({\text{-553.5kJmol}}^{\text{-1}}{\text{+0kJmol}}^{\text{-1}}\right)\text{-}\left({\text{0kJmol}}^{\text{-1}}\text{+}\left({\text{-58.5kJmol}}^{\text{-1}}\right)\text{+}\left({\text{-257.82kJmol}}^{\text{-1}}\right)\right)\\ {\text{=-553.5kJmol}}^{\text{-1}}\text{-}\left({\text{-295.68kJmol}}^{\text{-1}}\right)\\ {\text{=-257.82kJmol}}^{\text{-1}}\left({\text{1kJ=10}}^{\text{3}}\right)\\ {\text{=-257.82×10}}^{\text{3}}{\text{Jmol}}^{\text{-1}}\end{array}$

Hence,${\text{ΔG}}^{\text{o}}$for equation (a) is${\text{-257.82×10}}^{\text{3}}{\text{Jmol}}^{\text{-1}}$.

The standard potential.

Now,${\text{ΔE}}_{\text{cell}}^{\text{o}}$for equation (a) using equation (b);

Put; ${\text{ΔG}}^{\text{o}}{\text{=-257.82×10}}^{\text{3}}{\text{Jmol}}^{\text{-1}}$

n = 2,

F = 96485.34 C/ mol in equation (b),

role="math" localid="1663505420606" $\begin{array}{c}{\text{ΔE}}_{\text{cell}}^{\text{o}}\text{=}\frac{{\text{-257.85×10}}^{\text{3}}{\text{Jmol}}^{\text{-1}}}{{\text{-2×96485.34Cmol}}^{\text{-1}}}\\ \text{=1.34V}\end{array}$

Hence, the standard potential of zinc-mercury oxide cells is 1.34 V.