Question

Fundamentals of Electrolysis

17 - 6.The cell in Figure 17 - 4 is:

$\mathbf{Cu}\left|1.0{\mathrm{MCuSO}}_4\left(\mathrm{aq}\right)\right|\mathbf{KCL}\mathbf{(}\mathbf{aq}\mathbf{,}\mathbf{3}\mathbf{M}\mathbf{)}\left|\mathrm{AgCI}\left(s\right)\right|\mathbf{Ag}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$

Write half-reactions for this cell. Neglecting activity coefficients and the junction potential between${\mathbf{CuSO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$and KCI(aq), predict the equilibrium (zero-current) voltage expected when the Lugging capillary contacts the electrode. For this purpose, suppose that the reference electrode potential is 0.197Vvs. S.H.E. Why is the observed equilibrium potential+109mV, not the value you calculated?

How would the over potentials change if>1.000Vwere imposed by the

Potentiostat?

Short answer

(a) The Potential is E = 0.142 V

(b) The over potentials would increase because current is increased. The anode is intercepted when > 0.122 V , while the cathode when < 0.085 V

Step-by-step solution

Fundamentals of electrolysis

• Electrolysis is a fundamental process in chemistry that involves the breakdown of an electrolyte (a solution) and the creation of positive and negative ions.
• This is the underlying concept of electrolytic cells.
• The procedure may appear hard, yet it is as simple as a walk in the park.

Determine the potential

a)

The half-reactions for this cell is:

$$\begin{gathered} \mathrm{Cathode}:{\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to {\mathrm{CuSO}}_4 \\ \mathrm{Anode}:\mathrm{AGCL}+{\mathrm{e}}^{-}\to \mathrm{Aq}+{\mathrm{Cl}}^{-} \end{gathered}$$

The Potential is E = E ( cathode ) - E ( anode )

So, we first calculate E of cathode:

$$\begin{gathered} \mathrm{E}(\mathrm{cathode})={\mathrm{E}}^{°}-0.05916.\log \left(\left[{\mathrm{CuSO}}_4\right]\right) \\ \mathrm{E}(\mathrm{cathode})=0.339\mathrm{V}-0.339\mathrm{V} \\ \mathrm{E}(\mathrm{cathode})=0.339\mathrm{V} \end{gathered}$$

And, the E of reference electrode (anode) is:

$\mathrm{E}(\mathrm{cathode})=0.339\mathrm{V}$$\mathrm{E}(\mathrm{anode})=0.197\mathrm{V}$

The Potential is

$$\begin{gathered} \mathrm{E}=0.339\mathrm{V}-0.197\mathrm{V} \\ \mathrm{E}=0.142\mathrm{V} \end{gathered}$$

The predicted potential is 0.142V , while the observed potential is 0.109V is from ${\mathrm{Cu}}^{2+}$activity coefficient and using the 3MKCI instead of saturated KCI.

Step 2: Determine the over potential by the potentiostat

b)

If > 1.000V were imposed by the potentiostat, the over potentials would increase because current is increased.

The anode is intercepted when > 0.122V , while the cathode when .