7.2: Electrodeposition (Electrogravimetry)
- Page ID
- 81968
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Electrodeposition or electrogravimetry are two terms used to describe the same analysis method. The general procedure is to use something like a platinum electrode and apply a constant reducing potential that is sufficient to plate out a solid metal. For example, this method could be used to plate out cadmium metal from solution by the reaction shown below.
\[\mathrm{Cd^{2+}(aq) + 2e^- \rightleftharpoons Cd(s) \hspace{40px} E^o = - 0.403\: V}\]
The platinum is weighed before and after the plating step. The difference in weight is solid cadmium and this amount can be related back to the concentration of cadmium in the original solution.
Because it is essential that all the Cd2+ in solution get plated out onto the electrode, it is necessary use large surface area electrodes (on the order of 50 cm2) and to stir the solution to bring the Cd2+ up to the electrode. The plating step typically occurs at a potential well above the Eo value and over 30-60 minutes to insure that virtually all the Cd2+ has been plated onto the electrode. The plating potential must be well above the Eo value because, as the plating proceeds, the concentration of Cd2+ in solution will diminish, thereby raising the reducing potential needed for plating.
Will the presence of Fe2+ at 0.0800 M interfere with the electroplating of 99.9% of the Cd2+ in a solution in which the cadmium is expected to be present at a concentration of no less than 0.0500 M? Calculate potential values relative to a standard hydrogen electrode.
Solution
The first thing to consider is the two relevant half reactions for the metal ions in this problem.
\[\mathrm{Cd^{2+}(aq) + 2e^- \rightleftharpoons Cd(s) \hspace{40px} E^o = - 0.403\: V}\]
\[\mathrm{Fe^{2+}(aq) + 2e^- \rightleftharpoons Fe(s) \hspace{40px} E^o = - 0.440\: V}\]
Because the reduction potential of Cd2+ is less negative than that of Fe2+, it is easier to plate out the cadmium. This means that it may be possible to apply a reducing potential that is sufficient to plate out cadmium but not large enough to plate out any of the iron.
The next step is to focus on the plating of cadmium. We only need to consider the half reaction for cadmium for this calculation.
\[\mathrm{E= E^0 - \dfrac{0.059}{n} \log \dfrac{1}{[Cd^{2+}]}}\]
In this equation, n = 2. If we substitute in the initial concentration of Cd2+ (0.0500 M), we get the following value for E.
\[\mathrm{E= -0.403- \dfrac{0.059}{2} \log \dfrac{1}{0.0500} = -0.403-0.038= -0.441\: V}\]
Note, this is the potential needed to start the plating of cadmium. As the cadmium plates out, the concentration will be lower than 0.0500 M and that will impact the value of E. What we really need to consider is the concentration of Cd2+ when 99.9% of it has plated out. That would mean that only 0.1% of it is left in solution. 0.1% of 0.0500 M is 0.0000500 M (5.00 x 10-5 M). Let’s now reevaluate the potential for this solution.
\[\mathrm{E= -0.403- \dfrac{0.059}{2} \log \dfrac{1}{0.0000500}= -0.403-0.126= -0.529\: V}\]
So the potential needed to plate out 99.9% of the cadmium is –0.529 V. Note that as the concentration of Cd2+ drops, the potential needed to plate the cadmium becomes more negative (i.e,. the plating of cadmium becomes more difficult and requires a higher potential). Hopefully it seems reasonable that it would take a higher potential to force the cadmium to plate from solution as the concentration of Cd2+ becomes smaller.
The question we now need to answer is whether the potential needed to plate out 99.9% of the Cd2+ will begin to plate out the Fe2+. In this case, we only need to consider the potential that would be needed to start the plating of Fe2+, since plating of any iron will interfere with the measurement of cadmium. This requires using the half reaction for iron:
\[\mathrm{E= -0.440- \dfrac{0.059}{2} \log \dfrac{1}{0.0800}= -0.440-0.032= -0.472\: V}\]
Since the potential needed to begin plating out iron (– 0.472 V) is less negative than that needed to plate out 99.9% of the cadmium (– 0.529 V), it is not possible to selectively analyze for the cadmium in the presence of iron. The iron will plate out as well and interfere with the method.
Suppose the solution had 0.0800 M Cr3+ as a possible interference. Is it possible to plate out 99.9% of the cadmium without any interference from the chromium?
Solution
We first need to consider the appropriate half reaction for the reduction of Cr3+.
\[\mathrm{Cr^{3+}(aq) + 3e^- \rightleftharpoons Cr(s) \hspace{40px} E^o = - 0.744\: V}\]
Then we can solve for the potential that would just start the plating of chromium.
\[\mathrm{E= -0.744- \dfrac{0.059}{3} \log \dfrac{1}{0.0800}= -0.744-0.022= -0.766\: V}\]
Since the potential needed to begin plating out the chromium (– 0.766 V) is more negative than that needed to plate out 99.9% of the cadmium (– 0.529 V), it is possible to selectively plate out the cadmium. Setting the potential somewhere in the window between – 0.529 V and – 0.766 V relative to a standard hydrogen electrode would be necessary for the analysis.