# Why would hydrogen not be generated at the cathode when electrolysing zinc sulfate?

Describe how zinc metal can be obtained from zinc sulfate solution by electrolysis. A labelled diagram is acceptable. Include all the products of this electrolysis. The electrolysis is similar to that of copper (II) sulfate solution with inert electrodes.

The answer key says that zinc would be formed at the cathode. However, zinc is more reactive than hydrogen, so logically hydrogen should be given off at the cathode, not zinc.

As I understand it, the ions involved in this electrolysis are $\ce{Zn^2+}$, $\ce{H+}$, $\ce{OH-}$ and $\ce{SO4^2-}$.

Also, why can possible cathodes in this reaction be carbon, platinum OR zinc while the anode can only be carbon or platinum? What allows the cathode to be zinc?

In an electrolytic cell we are using the electrical energy from an outside power source to drive a chemical system against the direction it would spontaneously move. In this case, we can take a solution of zinc ions and force them to become zinc metal. As long as we put the zinc ions in an environment where they can be reduced, they will be.

Consider the half reaction that occurs when zinc metal is produced: $\ce{Zn^{2+}(aq) + 2e- -> Zn(s)}$

This is a reduction reaction, and reductions always occur at the cathode, whether the cell is galvanic or electrolytic.

The other half reaction involves hydroxide being converted into water and oxygen. $\ce{4 OH- -> 2 H2O + O2 + 4e-}$

This is an oxidation reaction, and oxidations always occur at the anode.

Platinum and carbon are nonreactive in this system, so they can be used for either electrode. Zinc can be used as a cathode because that's where it is being deposited, so we're just adding additional metallic zinc onto our already existing zinc electrode.

With the idea that oxidation occurs at the anode in mind, what would happen if we used a zinc anode?

• Zinc atoms would turn into zinc ions? But why would zinc ions be reduced at the cathode instead of hydrogen ions? That is the main issue I have. Oct 15 '14 at 2:45
• It's the concentration of H+ and especially overvoltage. Overvoltage is a fancy word for activation enthalpy. Oct 15 '14 at 3:08
• Um, can someone please reword this for someone taking his IGCSE's to understand? Oct 15 '14 at 4:04
• @dadadok Gotcha. In a basic solution, the concentration of H+ is very, very small. With a much larger concentration of zinc ions available and sufficient potential to make either reaction occur, the zinc reaction is far more likely by numbers alone. Both reactions ought to occur, though the H+ reaction will be very slow due to its low concentration. (Again, it's important that there is more than enough potential, the overvoltage Abel Friedman mentioned.) Oct 15 '14 at 4:14

You are correct that given the standard potentials in the table below, you would expect hydrogen to be reduced first under standard conditions.

\begin{array}{cc}\hline \text{Equation} & E_0 / \mathrm{V}\\ \hline \ce{Zn^2+ + 2 e- <=> Zn} & -0.7628\\ \ce{2 H+ + 2 e- <=> H2} & 0\\ \hline \end{array}

The key concept is standard conditions. Standard conditions assume that all relevant ions have a concentration of $1~\mathrm{mol/l}$ — and hydrogen is a relevant ion, thus standard conditions require $\mathrm{pH}\ 0$. To calculate the redox potential at non-standard conditions, the Nernst equation $(1)$ must be used. Thankfully, for the zinc equation it only depends on the concentration of zinc ions. I shall neglect zinc’s concentration for the remainder of this answer, because hydrogen is more interesting.

$$\begin{gather}E = E_0 - \frac{0.059~\mathrm{V}}{z} \lg \frac{[\ce{Red}]}{[\ce{Ox}]}\tag{1}\\[0.6em] E = 0~\mathrm{V}- \frac{7 \times 0.059~\mathrm{V}}{2} = -0.2065~\mathrm{V}\tag{2}\end{gather}$$

Equation $(2)$ tells us, that under neutral conditions, the potential of hydrogen is already $-0.21~\mathrm{V}$. This is still not enough to explain zinc formation at the cathode, but a big step forwards. More can be achieved by further raising of the $\mathrm{pH}$ value but it won’t get you all the way.

This is where a second issue comes in: overvoltage. Unfortunately, it cannot be expressed easily numerically (at least to the best of my knowledge) but in a nutshell the phenomenon is that it is hard to form gases on certain electrode materials (think activation energy). Due to the help of overvoltage (i.e. the correct choice of cathode), hydrogen liberaion can be almost completely suppressed and zinc formed.

Well, at reversible potentials (standard conditions), Hydrogen were to be discharged at the cathode but since it is a non-reversible type cell in which hydrogen Over-voltage is dependent on the combination of hydrogen atoms to form hydrogen molecules which is a slow process kinetically (RDS of the reaction). Hence, two factors affect the overvoltage phenomena. First, the nature of the cathode surface and second is the 'current density'.

It is seen that Platinised Platinum/palladium metal electrode act as a good catalyst for Hydrogen Evolution with lower values of over-voltage involved as compared to Lead/Mercury catalyst which not only are poor catalysts but also required high overvoltage for the reaction to occur.

Moreover, by plotting the Tafel Equation, this can be inferred that there is a parabolic relationship between over-voltage and current-density and hence as we increase the current density( upto a certain extent), deposition increases as we are supplying higher EMF as compared to what is offered by the deposited material as Back-EMF (when both become equal, deposition stops) and hence higher current density is needed for further deposition.

Also, after certain deposition of zinc metal on the cathode, hydrogen over-voltage on the zinc surface is of the order of 1 Volt. However, voltage required for Zinc deposition is 0.763V (Standard)/ non-ideally less than 1 Volt and hence, hydrogen ion will be discharged only when the zinc concentration in the solution becomes so small that the emf of zinc lies above the hydrogen overvoltage value.