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\documentclass[12pt]{article}
\title{Electrolysis of Water and Aqueous Solutions}
\author{Carlo Abelli \\ AP Chemistry --- Mr.\ Kern}
\date{February 19, 2015}
\usepackage[margin=1.0in]{geometry}
\begin{document}
\begin{titlepage}
\maketitle
\thispagestyle{empty}
\end{titlepage}
\section{Water Electrolysis Calculations and Questions}
\subsection{Half Reaction for Each Electrode}
Cathode (reduction): \(2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)}\)\\
This reaction took place at the cathode because during the splint test the gas
produced a popping sound, characteristic of hydrogen, a highly flammable gas.
\\
\\
Anode (oxidation): \(2H_{2}O_{(l)} \rightarrow O_{2(g)} + 4H^{+}_{(aq)} +
4e^{-}\)\\
This reaction took place at the anode because during the splint test the gas
reignited the splint ember, characteristic of oxygen.
\subsection{Overall Reaction}
\(2H_{2}O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)}\)
\subsection{Average Current}
Average Current: \(\frac{0.44 + 0.43 + 0.44 + 0.44 + 0.45 + 0.46 + 0.45 + 0.46
+ 0.46 + 0.45 + 0.45 + 0.46 + 0.45 + 0.47 + 0.46}{15} = 0.45\)
\subsection{Amount of Hydrogen Produced in One Hour}
\(\frac{0.45\ coulombs}{1.00\ \sec} * \frac{60.0\ \sec}{1.00\ \min} *
\frac{60.0\ \min}{1.00\ hour} * \frac{1.00\ e^{-}}{1.60*10^{-19}\ coulombs} *
\frac{1.00\ mol\ e^{-}}{6.022*10^{23}\ e^{-}} * \frac{1.00\ mol\ H_{2}}
{2.00\ mol\ e^{-}} = 0.0084\ mol\ H_{2}\)\\
\(0.0084\ mol\ H_{2} * 22.4\ L/mol = 0.19\ L\ H_{2}\)\\
\(0.0084\ mol\ H_{2} * 2.016\ g/mol = 0.0017\ g\ H_{2}\)
\subsection{Amount of Time Needed to Produce 30 mL Oxygen Gas at
22\(^{\circ}\)C and 1.00 atm}
\(1.00\ atm * 0.030\ L = n * 0.08206\ L\ atm\ K^{-1}\ mol^{-1} * 295\ K\)\\
\(n = 0.0012\ mol\ O_{2}\)\\
\(0.0012\ mol\ O_{2} * \frac{4.00\ mol\ e^{-}}{1.00\ mol\ O_{2}} *
\frac{6.022*10^{23}\ e^{-}}{1.00\ mol\ e^{-}} *
\frac{1.60*10^{-19}\ coulombs}{1\ e^{-}} * \frac{1.00\ \sec}{0.45\ coulombs}
= 100.\ \sec\)
\subsection{Moles of Gas Produced in Experiment}
Cathode (Hydrogen)\\
\(P_{container} = 36.8\ cm\ H_{2}O * \frac{10.0\ mm}{1.00\ cm} *
\frac{1.08\ g\ H_{2}O} {1\ mL\ H_{2}O} * \frac{1\ mL\ Hg}
{13.534\ g\ Hg} + 751.1\ mm\ Hg = 780.5\ mm\ Hg\)\\
\(P_{gas} = P_{container} - P_{H_{2}O}\)\\
\(P_{gas} = 780.5\ mm\ Hg - 18.7\ mm\ Hg = 761.8\ mm\ Hg\)\\
\(761.8\ mm\ Hg * 0.0486\ L = n * 62.363\ L\ mmHg\ K^{-1}\ mol^{-1} * 294.0\ K\)
\\
\(n = 0.00202\ mol\ H_{2}\)\\
\\
Anode (Oxygen)\\
\(P_{container} = 24.0\ cm\ H_{2}O * \frac{10.0\ mm}{1.00\ cm} *
\frac{1.08\ g\ H_{2}O} {1\ mL\ H_{2}O} * \frac{1\ mL\ Hg}
{13.534\ g\ Hg} + 751.1\ mm\ Hg = 770.3\ mm\ Hg\)\\
\(P_{gas} = P_{container} - P_{H_{2}O}\)\\
\(P_{gas} = 770.3\ mm\ Hg - 18.7\ mm\ Hg = 751.6\ mm\ Hg\)\\
\(751.6\ mm\ Hg * 0.0239\ L = n * 62.363\ L\ mmHg\ K^{-1}\ mol^{-1} * 294.0\ K\)
\\
\(n = 0.000980\ mol\ O_{2}\)
\subsection{Theoretical Moles of Gas Produced}
Cathode (Hydrogen)\\
\(\frac{0.45\ coulombs}{1.00\ \sec} * \frac{60.0\ \sec}{1.00\ \min} * 14.0\ \min *
\frac{1.00\ e^{-}}{1.60*10^{-19}\ coulombs} * \frac{1.00\ mol\ e^{-}}
{6.022*10^{23}\ e^{-}} * \frac{1.00\ mol\ H_{2}}{2.00\ mol\ e^{-}} =
0.0020\ mol\ H_{2}\)\\
\\
Anode (Oxygen)\\
\(\frac{0.45\ coulombs}{1.00\ \sec} * \frac{60.0\ \sec}{1.00\ \min} * 14.0\ \min *
\frac{1.00\ e^{-}}{1.60*10^{-19}\ coulombs} * \frac{1.00\ mol\ e^{-}}
{6.022*10^{23}\ e^{-}} * \frac{1.00\ mol\ O_{2}}{4.00\ mol\ e^{-}} =
0.00098\ mol\ O_{2}\)
\subsection{Percent Deviation}
Cathode (Hydrogen)\\
\(\frac{0.00202\ mol - 0.0020\ mol}{0.0020\ mol} * 100 = 0.0\%\ deviation\)\\
\\
Anode (Oxygen)\\
\(\frac{0.000980\ mol - 0.00098\ mol}{0.00098\ mol} * 100 = 0.0\%\ deviation\)
\section{Aqueous KI, KBr and KCl Electrolysis Calculations and Questions}
\subsection{Half Reactions for Each Electrode}
\begin{tabular}{ l r @{\(\rightarrow\)} l }
KI Solution\\
Cathode (reduction): & \(2H_{2}O_{(l)} + 2e^{-}\) &
\(2OH^{-}_{(aq)} + H_{2(g)}\)\\
Anode (oxidation): & \(2I^{-}_{(aq)}\) & \(I_{2(s)} + 2e^{-}\)\\
\\
KBr Solution\\
Cathode (reduction): & \(2H_{2}O_{(l)} + 2e^{-}\) &
\(2OH^{-}_{(aq)} + H_{2(g)}\)\\
Anode (oxidation): & \(2Br^{-}_{(aq)}\) & \(Br_{2(l)} + 2e^{-}\)\\
\\
KCl Solution\\
Cathode (reduction): & \(2H_{2}O_{(l)} + 2e^{-}\) &
\(2OH^{-}_{(aq)} + H_{2(g)}\)\\
Anode (oxidation): & \(2Cl^{-}_{(aq)}\) & \(Cl_{2(g)} + 2e^{-}\)\\
\end{tabular}\\
\\
The water half reaction had to have occurred at the cathode because the
phenolphthalein indicator was activated by the creation of \(OH^{-}\),
creating the purple color observed at the cathode.\\
\\
The iodine, chlorine and bromine half reactions had to have occurred at the
anode because in the first experiment, a dark solid was created at the anode
(iodine), in the second experiment, a yellow liquid was observed (bromine)
and in the third experiment, a chlorine smelling gas was produced (chlorine).
\subsection{Overall Equations}
\begin{tabular}{ l r @{\(\rightarrow\)} l }
KI Solution: & \(KI_{(aq)} + 2H_{2}O_{(l)}\) & \(KOH_{(aq)} + 2H_{2(g)} +
I_{2(s)}\)\\
KBr Solution: & \(KBr_{(aq)} + 2H_{2}O_{(l)}\) & \(KOH_{(aq)} + 2H_{2(g)} +
Br_{2(l)}\)\\
KCl Solution: & \(KCl_{(aq)} + 2H_{2}O_{(l)}\) & \(KOH_{(aq)} + 2H_{2(g)} +
Cl_{2(g)}\)\\
Test Solution:
\end{tabular}
\subsection{Half Reactions and Overall Equation for Molten KI, KBr and KCl}
KI\\
\begin{tabular}{ l r @{\(\rightarrow\)} l }
Cathode (reduction): & \(K^{+}_{(l)} + e^{-}\) & \(K_{(s)}\)\\
Anode (oxidation): & \(2I^{-}_{(l)}\) & \(I_{2(s)} + 2e^{-}\)\\
Overall Equation: & \(2KI_{(l)}\) & \(2K_{(s)} + I_{2(s)}\)\\
\end{tabular}\\
\\
KBr\\
\begin{tabular}{ l r @{\(\rightarrow\)} l }
Cathode (reduction): & \(K^{+}_{(l)} + e^{-}\) & \(K_{(s)}\)\\
Anode (oxidation): & \(2Br^{-}_{(l)}\) & \(Br_{2(l)} + 2e^{-}\)\\
Overall Equation: & \(2KBr_{(l)}\) & \(2K_{(s)} + Br_{2(s)}\)\\
\end{tabular}\\
\\
KCl\\
\begin{tabular}{ l r @{\(\rightarrow\)} l }
Cathode (reduction): & \(K^{+}_{(l)} + e^{-}\) & \(K_{(s)}\)\\
Anode (oxidation): & \(2Cl^{-}_{(l)}\) & \(Cl_{2(g)} + 2e^{-}\)\\
Overall Equation: & \(2KCl_{(l)}\) & \(2K_{(s)} + Cl_{2(g)}\)\\
\end{tabular}
\subsection{Time Required to Produce 1.25 g Iodine at 0.75 amps}
\(1.25\ g\ I_{2} * \frac{1.00\ mol\ I_{2}}{253.80\ g} *
\frac{2.00\ mol\ e^{-}}{1.00\ mol\ I_{2}} *
\frac{6.022*10^{23}\ e^{-}}{1.00\ mol\ e^{-}} *
\frac{1.60*10^{-19}\ coulombs}{1.00\ e^{-}} *
\frac{1.00\ \sec}{0.75\ coulombs} *
\frac{1.00\ \min}{60.0\ \sec} = 21\ \min\)
\subsection{Time Required to Produce 0.100 L Hydrogen gas in Lab Conditions at
0.75 amps}
\(751.1\ mmHg * 0.100\ L\ H_{2} = n * 62.363\ L\ mmHg\ K^{-1}\ mol^{-1} *
294\ K\)\\
\(n = 0.00410\ mol\)\\
\(0.00410\ mol\ H_{2} * \frac{2.00\ mol\ e^{-}}{1.00\ mol\ H_{2}} *
\frac{6.022*10^{23}\ e^{-}}{1.00\ mol\ e^{-}} *
\frac{1.60*10^{-19}\ coulombs}{1.00\ e^{-}} *
\frac{1.00\ \sec}{0.75\ coulombs} * \frac{1.00\ \min}{60.0\ \sec} = 18\ \min\)
\end{document}
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¿Por qué está pasando esto? No puedo resolverlo. Ejecutar pdflatexdesde la línea de comando genera el pdf (después de generar algunas advertencias), por lo que el problema debe provenir de o latexmk, ¿verdad?vimtexvim
Respuesta1
Se arregló usando \usepackage{fixltx2e}para hacer \(y \)robusto. Ahora compila sin errores.