Electrochemistry
- Page ID
- 283142
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)CHEM 2300, In-class Exercises, Class 25
Name: _________________________
- Write the relevant half reactions and calculate Eo and K for the following reaction. Is the reaction spontaneous as written?
\[\ce{I2(s) + 5 Br2(aq) + 6 H2O ↔ 2IO3- + 10Br- + 12H+}\nonumber\]
\[\mathrm{E^o_{I_2} = 1.210}\nonumber\]
\[\mathrm{E^o_{Br_2} = 1.098}\nonumber\]
- Write the line notation for a cell made up of the following half reactions if the indicator electrodes are platinum and a monoprotic acid is in solution in the cathode cell. If a balanced reaction is not given, assume that the larger Eo value is the cathode.
\[\ce{Fe^3+ + e- ↔ Fe^2+} \hspace{50px} \mathrm{E^0 = 0.771}\nonumber\]
\[\ce{Cr2O7^2- + 14H+ +6e- ↔ 2Cr^3+ + 7H2O} \hspace{30px} \mathrm{E^0 = 1.36}\nonumber\]
- Given the following reaction, assign oxidation states to each element and write the 2 half reactions (both as reductions).
\[\ce{2 AgCl(s) + Pb(s) + 2F- (aq) ↔ 2 Ag(s) + 2Cl- (aq) + PbF2(s)}\nonumber\]
- Write the line notation for this cell.
- If the concentrations of NaF and KCl are both 0.10 M, calculate the cell potential for the cell made up of the above half reactions at 25 oC:
CHEM 2300, In-class Exercises, Class 26
Name: _________________________
- The apparatus in the figure below was used to monitor the titration of 50.0 mL of 0.100 M AgNO3 with 0.200 M NaBr. Calculate the cell voltage at each volume of NaBr, below.
\[\mathrm{K_{sp}\: for\: AgBr(s) = 5.0 \times 10^{-13}}\nonumber\]
\[\mathrm{E_{S.C.E.}=0.241\: V}\nonumber\]
- 1.00 mL
- 25.1
- Write the half reaction(s) for the following reactions, as reductions:
- Pt (s) | Cr3+ (0.10 M), Cr2+ (0.050 M)
- Cd(s) | Cd2+ (1 M) || H+(aq, 1 M), Mn2+ (l M), MnO4- (l M) | Pt
- Cr2O72-(aq), Cr3+(aq), HA(aq) |Pt(s)
- Pt (s) | Cr3+ (0.10 M), Cr2+ (0.050 M)
- A solution contains 0.100 M Ce3+, 1.00x10-4 M Ce4+, 1.00x10-4 M Mn2+, 0.100 M MnO4-, and 1.00M HClO4.
- Write a balanced net reaction that can occur between species in this solution, assuming the Ce half cell is the cathode.
- If Ecell is measured at -0.019 V, calculate the equilibrium constant for the balanced net reaction.
\[\mathrm{K=10^{n*E^o/0.05916}}\nonumber\]
Contributors and Attributions
- Sarah Gray, Stockton University (sarah.gray@stockton.du)
- Sourced from the Analytical Sciences Digital Library