Module 6

Lesson 3.3  Half-Cell Stoichiometry



Key Concepts


In the previous section, you learned how to calculate the number of moles of electrons transferred in a given redox reaction. Recall that a balanced chemical reaction (including a half-reaction) provides information about the proportions of substances in a chemical process. With knowledge of the quantity of electrons, you are able to determine the quantities of other species in the half-reaction.


Whether you are performing stoichiometric calculations for a voltaic cell or an electrolytic cell, the basic steps are the same. Because the mass of an element produced at an electrode depends on the amount of transferred electrons, you first need a balanced half-reaction showing the moles of electrons involved.


For example, when zinc is plated onto a steel pipe, two moles of electrons must be gained by one mole of zinc to deposit one mole of zinc atoms as metal.


\( \mathrm { Zn^{2+} + 2e^- \rightarrow Zn(s) } \)


We cannot measure the number of electrons directly, but we can indirectly determine the number of moles of electrons transferred in a chemical reaction by employing Faraday's Law. After the number of moles of transferred electrons are calculated, the mole ratios displayed in the balanced half-reactions can be used to calculate the mass of substances consumed or produced at the electrodes.


Study the two examples provided below.



Example I (Voltaic Cell)


For an example of voltaic cell stoichiometry, consider Section 14.4 Question 8 from page 657 of the textbook.


"A student reconstructs Volta's electric battery using sheets of copper and zinc, and a current of 0.500 A is produced for 10.0 minutes. Calculate the mass of zinc oxidized to aqueous zinc ions."


Step One: Write a half-reaction for the oxidation of zinc solid to zinc ions.


\( \mathrm { Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- } \)


Step Two: Calculate the moles of electrons transferred using Faraday's Law.


\({ n_\mathrm{e^-} = \dfrac{It}{F} } \) 


\(n_\mathrm{e^-} = \mathrm{\dfrac{ 0.500~A \times (10.0~min \times 60~\frac{s}{min}) }{ 9.65 \times 10^4 ~\frac{C}{s} } } \)


\(  n_\mathrm{e^-} = \mathrm{0.00311~ mol } \)


Step Three: Calculate the moles of zinc solid using the mole ratio from the half-reaction.


\( n_\mathrm{zn} = x \dfrac{1}{2} = \mathrm{0.00155 ~mol } \)


Step Four: Calculate mass of zinc solid.


\(  { m_\mathrm{zn} = n_\mathrm{zn}\times M } \)


\( m_\mathrm{zn} = \mathrm{0.00155 ~mol \times 65.41~\frac{g}{mol} = 0.102~ g } \)



Example II (Electrolytic Cell)


For an example of electrolytic cell stoichiometry, consider the following problem.


Nickel provides a decorative appearance due to its ability to cover imperfections in the base metal. Calculate the mass of nickel plated if a current of 1.85 A is applied to an electrochemical cell that contains a solution of nickel(II) sulfate, nickel(II) chloride, and boric acid for 60.0 h.


Step One: Write a reduction half-reaction for the nickel-plating process.


\( \mathrm { Ni^{2+}(aq) + 2e^- \rightarrow Ni(s) } \)


Step Two: Calculate the moles of electrons transferred using Faraday's Law.


\( { n_\mathrm{e^-} = \dfrac{It}{F} } \)


\( n_\mathrm{e^-} = \mathrm {\dfrac{ 1.85~A \times (60.0~h \times 3600~\frac{s}{h}) }{ 9.65 \times 10^4~\frac{C}{s} } } \)


\( n_\mathrm{e^-} = \mathrm{4.14~ mol } \)


Step Three: Calculate moles of nickel using the mole ratio from the half-reaction.


\( { n_\mathrm{Ni} = \mathrm{4.14~mol \times \dfrac{1}{2} = 2.07~ mol }} \)


Step Four: Calculate mass of nickel.

\( m_\mathrm{Ni} = n \times M = \mathrm{2.07~ mol \times 58.69 \frac{g}{mol} = 122~ g } \)



    Read pages 655-656 in your textbook.

Check Your Understanding


Complete Section 14.4 Questions 2, 3, 9, 11, and 12 on page 657 of the textbook.


Click the link below to check your work

Page 657 Section 14.4 Question 2


\( \mathrm { 2 Cl^-(aq) \rightarrow Cl_2(g) + 2e^- } \)


\( \mathrm { t = (8.0 ~h) \times \dfrac{3600~ s}{1~ h} = 2.9 \times 10^4 ~s } \)


\( \mathrm { n_e = \dfrac{It}{F} } \)


\( \mathrm { = \dfrac{(5.5 \times 10^4 ~\frac{C}{s})(2.9 \times 10^4~ s)}{9.65 \times 10^4~\frac{C}{mol}} } \)


\( \mathrm { = 1.6 \times 10^4~ mol } \)


\( \mathrm { n_{Cl_2} = 1.6 \times 10^4 ~mol \times \dfrac{1}{2} = 8.2 \times 10^3~ mol } \)


\( \mathrm { m_{Cl_2} = (8.2 \times 10^3 ~mol)(70.90 ~\frac{g}{mol}) = 5.8 \times 10^5 ~g } \)


The mass of chlorine formed is 5.8 x 105 g, or 5.8 x 102 kg.



Page 657 Section 14.4 Question 3


\( \mathrm { Ag^+(aq) + 1e^- \rightarrow Ag(s) } \)


\( \mathrm { n_{Ag} = \dfrac{10.00~ g}{107.87~\frac{g}{mol}} = 0.09270~ mol } \)


\( \mathrm { n_e = 0.09270~mol \times \dfrac{1}{1} = 0.09270~ mol } \)


\( \mathrm { t = \dfrac{n_eF}{I} } \)


\( \mathrm { = \dfrac{(0.09270~ mol)(9.65 \times 10^4 ~\frac{C}{mol})}{(1.80~\frac{C}{s})} } \)


\( \mathrm { = 4.97 \times 10^3 ~s } \)


\( \mathrm { t = (4.97 \times 10^3 s) \times \dfrac{1~ min}{60~s} = 82.8~min } \)



Page 657 Section 14.4 Question 9

  1. \( \mathrm { Cr^{3+}(aq) + 3e^- \rightarrow Cr(s) } \)


    \( \mathrm { t = 45.5~min \times \dfrac{60~s}{1~min} = 2.73 \times 10^3~ s } \)


    \( \mathrm { n_e = \dfrac{It}{F} = \dfrac{(54~\frac{C}{s})(2.73 \times 10^3~ s)}{(9.65 \times 10^4~ \frac{C}{mol})} = 1.53~mol } \)


    \( \mathrm { n_{Cr} = 1.53~mol \times \dfrac{1}{3} = 0.509~mol } \)


    \( \mathrm { m_{Cr} = (0.509~ mol)(52.00 ~\frac{g}{mol}) = 26~ g } \)


    The mass of chromium deposited is 26 g.


  2. \( \mathrm { Ni^{2+}(aq) + 2e^- \rightarrow Ni(s) } \)


    \( \mathrm { n_{Ni} = \dfrac{0.250~ g}{58.69 ~\frac{g}{mol}} = 0.00426~mol } \)


    \( \mathrm { n_e = 0.00426 ~mol \times \dfrac{2}{1} = 0.00852~mol } \)


    \( \mathrm { t = \dfrac{n_eF}{I} = \dfrac{(0.00852 ~mol)(9.65 \times 10^4 ~\frac{C}{mol})}{0.540~\frac{C}{s}} } \)


    \( \mathrm { = 1.52 \times 10^3 ~s } \)


    \( \mathrm { t = 1.52 \times 10^3~ s \times \dfrac{1~ min}{60~ s} = 25.4~ min } \)


    The time required is 25.4 min.



Page 657 Section 14.4 Question 11


\( \mathrm { Cr_2O_7^{2-}(aq) \rightarrow Cr(s) } \)


\( \mathrm { Cr_2O_7^{2-}(aq) \rightarrow 2 Cr(s) } \)


\( \mathrm { Cr_2O_7^{2-}(aq) \rightarrow 2 Cr(s) + 7 H_2O(l) } \)


\( \mathrm { Cr_2O_7^{2-}(aq) + 14 H^+(aq) \rightarrow 2 Cr(s) + 7 H_2O(l) } \)


\( \mathrm { Cr_2O_7^{2-}(aq) + 14 H^+(aq) + 12 e^- \rightarrow 2 Cr(s) + 7 H_2O(l) } \)


\( \mathrm { n_{Cr} = \dfrac{17.8~g}{52.00~\frac{g}{mol}} = 0.342~ mol } \)


\( \mathrm { n_e = 0.342~ mol \times \dfrac{12}{2} = 2.05~ mol } \)


\( \mathrm { t = 2.20~ h = 2.20~ h \times \dfrac{3600~s}{1~ h} = 7.92 \times 10^3~ s } \)


\( \mathrm { I = \dfrac{n_eF}{t} = \dfrac{(2.05 ~mol)(9.65 \times 10^4 \frac{C}{mol})}{(7.92 \times 10^3~ s)} = 25.0 ~\frac{C}{s} } \)


The average current required is 25.0 C/s.



Page 657 Section 14.4 Question 12


Prediction


\( \mathrm { Sn^{2+}(aq) + 2e^- \rightarrow Sn(s) } \)


\( \mathrm { t = 6.00~min \times \dfrac{60~s}{1~min} = 360~ s } \)


\( \mathrm { n_e = \dfrac{It}{F} = \dfrac{(3.46~\frac{C}{s})(360~ s)}{9.65 \times 10^4} = 0.01291~mol } \)


\( \mathrm { n_{Sn} = 0.01291~ mol \times \dfrac{1}{2} = 0.00645 ~mol } \)


\( \mathrm { m_{Sn} = (0.00645 ~mol)(118.71~ \frac{g}{mol}) = 0.766 ~g } \)


The mass of tin plated onto the can will be 0.766 g.



Analysis

m(Sn) deposited on can = 118.05 g - 117.34 g = 0.71 g

The mass of tin electroplated onto the can was 0.71 g.


Evaluation

\( \mathrm { \%_{difference} = \dfrac{|(0.71~g - 0.766~g)|}{0.766~g} \times 100 } \)


\( \mathrm { = 7.3 \% } \)


The prediction is judged to be verified. A response that the result is inconclusive is acceptable also.