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class="mark-page-favorite pull-right" data-pid="907" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../chemistry.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../362/redox-processes.html">Redox Processes</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Electrochemical cells II</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 60 minutes"><i class="fa fa-clock-o"></i> 60&apos;</span> </ol> <article id="main-article"> <p><img alt="" src="../../images/redox-processes/danielcell.jpg" style="width: 160px; height: 148px; float: left;">It is absolutely critical that understanding of the topics <em>oxidation and reduction 9.1</em> and <em>electrochemical cells 9.2 </em>is thorough before beginning on revision of this, more advanced, topic. It is so easy to get lost in this topic, and there is a lot to deal with. However, a really sound grasp of the earlier topics; knowing exactly how voltaic and electrolytic cells operate and where oxidation and reduction occurs will make this much easier, as we can always return to first principles and ask ourselves, <em>Where are the electrons moving?</em></p> <hr class="hidden-separator"> <div class="panel panel-has-colored-body panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key concepts</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <p>Ensure you are confident using the terms below and learn the asterisked* definitions</p> <p>Electrochemical series, <strong>Standard electrode potential*</strong>, Cell potential, <strong>Standard hydrogen electrode (SHE)*</strong>, Faraday constant, Electroplating, Absolute charge</p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="3" data-mode="Normal" data-topics="645" data-subject-id="7" data-n-flashcards="7" style="text-align:center">Show flashcards</a></div><hr> <p>&nbsp;</p> <div class="panel panel-blue"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Electrochemical cells</p> </div> </div> <div class="panel-body"> <div> <p>A reminder from <em>electrochemical cells</em> <em>9.1</em> that understanding the redox processes that take place in voltaic and electrolytic cells is a key concept.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/447161708"></iframe></div> <p>&nbsp;</p> </div> </div> <div class="panel-footer"> <div> <p>&nbsp;</p> </div> </div> </div> </div> </div> </div> </div> <div class="panel-footer"> <div>&nbsp;</div> </div> </div> <div class="panel panel-yellow 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panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test yourself</p> </div> </div> <div class="panel-body"> <div> <div class="panel-body"> <div> <div class="tib-quiz" data-stats="7-538-907"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following statements about a voltaic cell consisting of the zinc and iron half-cells below are correct?</p><p>Zn<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ Zn<sub>(s) </sub> E<sup><s>o</s></sup> = &ndash;0.76 V</p><p>Fe<sup>3+</sup><sub>(aq)</sub> + e<sup>&ndash;</sup> ⇌ Fe<sup>2+</sup><sub>(aq) </sub> E<sup><s>o</s> </sup>= +0.77 V</p><p><strong>1: </strong>Electrons flow from the iron half-cell to the zinc half-cell.</p><p><strong>2: </strong>The cell potential E<sup><s>o</s></sup><sub>cell</sub> = +1.53 V</p><p><strong>3: </strong>Negative ions travel across the salt bridge from the iron half-cell to the zinc half-cell.</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1, 2 and 3</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 3 only</span></label> </p></div><div class="q-explanation"><p>In a voltaic cell the redox reaction generates the charges on the electrodes and causes the current to flow. The half-cell highest in the electrochemical series (most negative) loses electrons (oxidation) causing the electrode to be negative, and the half-cell lowest in the electrochemical series (most positive) gains electrons (reduction) causing the electrode to be positive.</p><p>Zinc half-cell is more negative than the iron half-cell (Electrochemical series in the data book) so the zinc half-cell will lose electrons (oxidation) and the iron half-cell will gain electrons (reduction); <strong>electrons will flow from the zinc to iron</strong>.</p><p>E<sup><s>o</s></sup>cell = E<sup><s>o</s></sup>cathode &ndash; E<sup><s>o</s></sup>anode = +0.77 &ndash; &ndash;0.76 = <strong>+1.53 V</strong></p><p>As electrons are flowing from zinc to iron, the charge will be equalised by n<strong>egative ions flowing across the salt bridge from the iron to the zinc </strong>half-cell.</p><p>Thus <strong>2 and 3 only </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the overall cell reaction and cell potential for a voltaic cell consisting of the nickel and copper half-cells below?</p><p>Ni<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ Ni<sub>(s) </sub> E<sup><s>o</s></sup> = &ndash;0.26 V</p><p>Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ Cu<sub>(s) </sub> E<sup><s>o</s></sup> = +0.34 V</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Cu<sup>2+</sup><sub>(aq)</sub> + Ni<sub>(s)</sub> ⇌ Ni<sup>2+</sup><sub>(aq)</sub> + Cu<sub>(s)</sub> E<sup><s>o</s></sup>cell = +0.08 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>Ni<sup>2+</sup><sub>(aq)</sub> + Cu<sub>(s)</sub> ⇌ Cu<sup>2+</sup><sub>(aq)</sub> + Ni<sub>(s)</sub> E<sup><s>o</s></sup>cell = +0.08 V</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Cu<sup>2+</sup><sub>(aq)</sub> + Ni<sub>(s)</sub> ⇌ Ni<sup>2+</sup><sub>(aq)</sub> + Cu<sub>(s)</sub> E<sup><s>o</s></sup>cell = +0.60 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>Ni<sup>2+</sup><sub>(aq)</sub> + Cu<sub>(s)</sub> ⇌ Cu<sup>2+</sup><sub>(aq)</sub> + Ni<sub>(s)</sub> E<sup><s>o</s></sup>cell = +0.60 V</span></label> </p></div><div class="q-explanation"><p>In a voltaic cell the half-cell highest in the electrochemical series (most negative) loses electrons (oxidation) and the half-cell lowest in the electrochemical series (most positive) gains electrons (reduction).</p><p>Nickel half-cell is more negative than the copper half-cell (Electrochemical series in the data book) so the nickel half-cell will lose electrons (oxidation) and the copper half-cell will gain electrons (reduction)<strong>:</strong></p><p>Ni<sub>(s)</sub> &rarr; Ni<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> and Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> &rarr; Cu<sub>(s)</sub></p><p>The number of electrons lost and gained is the same so the half-equations combined:</p><p>Cu<sup>2+</sup><sub>(aq)</sub> + Ni<sub>(s)</sub> ⇌ Ni<sup>2+</sup><sub>(aq)</sub> + Cu<sub>(s)</sub></p><p>E<sup><s>o</s></sup>cell = E<sup><s>o</s></sup>cathode &ndash; E<sup><s>o</s></sup>anode = +0.34 &ndash; &ndash;0.26 = +0.60 V</p><p>Thus <strong>Cu<sup>2+</sup><sub>(aq)</sub> + Ni<sub>(s)</sub> ⇌ Ni<sup>2+</sup><sub>(aq)</sub> + Cu<sub>(s)</sub> E<sup><s>o</s></sup>cell = +0.60 V</strong> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the overall cell reaction and cell potential for a voltaic cell consisting of the nickel and aluminium half-cells below?</p><p>Ni<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ Ni<sub>(s) </sub> E<sup><s>o</s></sup> = &ndash;0.26 V</p><p>Al<sup>3+</sup><sub>(aq)</sub> + 3e<sup>&ndash;</sup> ⇌ Al<sub>(s) </sub> E<sup><s>o</s></sup> = &ndash;1.66 V</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2Al<sup>3+</sup><sub>(aq)</sub> + 3Ni<sub>(s) </sub>⇌ 3Ni<sup>2+</sup><sub>(aq)</sub> + 2Al<sub>(s)</sub> E<sup><s>o</s></sup>cell = &ndash;1.40 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>2Al<sup>3+</sup><sub>(aq)</sub> + 3Ni<sub>(s) </sub>⇌ 3Ni<sup>2+</sup><sub>(aq)</sub> + 2Al<sub>(s)</sub> E<sup><s>o</s></sup>cell = &ndash;1.92 V</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>3Ni<sup>2+</sup><sub>(aq)</sub> + 2Al<sub>(s)</sub> ⇌ 2Al<sup>3+</sup><sub>(aq)</sub> + 3Ni<sub>(s)</sub> E<sup><s>o</s></sup>cell = +1.40 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>Ni<sup>2+</sup><sub>(aq)</sub> + Al<sub>(s)</sub> ⇌ Al<sup>3+</sup><sub>(aq)</sub> + Ni<sub>(s)</sub> E<sup><s>o</s></sup>cell = +1.40 V</span></label> </p></div><div class="q-explanation"><p>In a voltaic cell the half-cell highest in the electrochemical series (most negative) loses electrons (oxidation) and the half-cell lowest in the electrochemical series (most positive) gains electrons (reduction).</p><p>Aluminium half-cell is more negative than the nickel half-cell (Electrochemical series in the data book) so the aluminium half-cell will lose electrons (oxidation) and the nickel half-cell will gain electrons (reduction)<strong>:</strong></p><p>Al<sub>(s)</sub> &rarr; Al<sup>3+</sup><sub>(aq)</sub> + 3e<sup>&ndash;</sup> and Ni<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> &rarr; Ni<sub>(s)</sub></p><p>The number of electrons lost and gained in each half-cell must be equalised when the half-equations combine (Al &times; 2, Ni &times; 3):</p><p>3Ni<sup>2+</sup><sub>(aq)</sub> + 2Al<sub>(s)</sub> ⇌ 2Al<sup>3+</sup><sub>(aq)</sub> + 3Ni<sub>(s)</sub></p><p>E<sup><s>o</s></sup>cell = E<sup><s>o</s></sup>cathode &ndash; E<sup><s>o</s></sup>anode = &ndash;0.26 &ndash; &ndash;1.66 = +1.40 V</p><p>Note that the E<sup><s>o</s></sup> values are independent of the multiplication to equal out the number of electrons.</p><p>Thus <strong>3Ni<sup>2+</sup><sub>(aq)</sub> + 2Al<sub>(s)</sub> ⇌ 2Al<sup>3+</sup><sub>(aq)</sub> + 3Ni<sub>(s)</sub> E<sup><s>o</s></sup>cell = +1.40V</strong> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following statements about the standard hydrogen electrode (SHE) are correct?</p><p><strong>1: </strong>The electrode is made of inert platinum.</p><p><strong>2: </strong>Hydrogen gas is fed onto the electrode at 100kPa.</p><p><strong>3: </strong>The electrode is immersed in hydrochloric acid (1 mol dm<sup>&minus;3</sup>) at 298K.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2 and 3 only</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1, 2 and 3</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 3 only</span></label> </p></div><div class="q-explanation"><p>Electrode potentials are measured against the SHE; standard hydrogen electrode.</p><p>The SHE consists of a feed of hydrogen gas at 100 kPa onto an inert Pt electrode in 1 mol dm<sup>&ndash;3</sup> hydrochloric acid all at 298 K.</p><p>The platinum electrode is coated with platinum black; finely powdered platinum to give a high surface area for the reaction to occur on: H<sup>+</sup><sub>(aq)</sub> + e<sup>&ndash;</sup> ⇌ &frac12;H<sub>2(g)</sub> E<sup><s>o</s></sup> = 0.00 V</p><p>Thus <strong>1, 2 and 3 </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which of the following half-cells <strong>does not </strong>require a platinum electrode?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>Cu<sup>2+</sup><sub>(aq)</sub> + e<sup>&ndash;</sup> ⇌ Cu<sup>+</sup><sub>(aq)</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ Cu<sub>(s)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span>H<sup>+</sup><sub>(aq)</sub> + e<sup>&ndash;</sup> ⇌ &frac12;H<sub>2(g)</sub></span></label> </p><p><label class="radio"> <input type="radio"> <span> &frac12;Cl<sub>2(g)</sub> + e<sup>&ndash;</sup> ⇌ Cl<sup>&minus;</sup><sub>(aq)</sub></span></label> </p></div><div class="q-explanation"><p>An inert platinum electrode is needed for all gaseous electrodes (e.g. chlorine and hydrogen) and for half-cells involving only aqueous ions (e.g. Cu<sup>2+</sup><sub>(aq)</sub> + e<sup>&ndash;</sup> ⇌ Cu<sup>+</sup><sub>(aq)</sub>).</p><p>The copper half-cell <strong>Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ Cu<sub>(s)</sub> </strong>requires a copper electrode and is therefore the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Calculator question: </em>What is the value of &Delta;G<sup><s>o</s></sup> under standard conditions when calculated for the cell reaction in a voltaic cell consisting of the two half-cells below (in kJ to 3 sig figs)?</p><p>(using &Delta;G<sup><s>o</s></sup>= &ndash;nFE<sup><s>o</s></sup><sub>cell</sub> in data book)</p><p>Fe<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ Fe<sub>(s) </sub> E<sup><s>o</s></sup> = &ndash;0.45 V</p><p>Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ Cu<sub>(s) </sub> E<sup><s>o</s></sup> = +0.34 V</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>&ndash;152 </span></label> </p><p><label class="radio"> <input type="radio"> <span>+76.2</span></label> </p><p><label class="radio"> <input type="radio"> <span>&ndash;76.2</span></label> </p><p><label class="radio"> <input type="radio"> <span>+152</span></label> </p></div><div class="q-explanation"><p>Iron(II) half-cell is more negative than the copper half-cell (Electrochemical series in the data book) so the iron half-cell will lose electrons (oxidation) and the copper half-cell will gain electrons (reduction)<strong>:</strong></p><p>Fe<sub>(s)</sub> &rarr; Fe<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> and Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> &rarr; Cu<sub>(s)</sub></p><p>The number of electrons lost and gained is the same (<strong>that is two electrons</strong>) so the half-equations combined:</p><p>Cu<sup>2+</sup><sub>(aq)</sub> + Fe<sub>(s)</sub> ⇌ Fe<sup>2+</sup><sub>(aq)</sub> + Cu<sub>(s)</sub></p><p>E<sup><s>o</s></sup>cell = E<sup><s>o</s></sup>cathode &ndash; E<sup><s>o</s></sup>anode = +0.34 &ndash; &ndash;0.45 = +0.79 V</p><p>&Delta;G<sup><s>o</s></sup> = &ndash;nFE<sup><s>o</s></sup><sub>cell </sub> = &ndash; (<strong>2 mol</strong> &times; 96500 C mol<sup>&ndash;1</sup> &times; +0.79 V) = &ndash; 152470 J = &ndash;152 kJ (3 sig figs)</p><p><strong>Incorrect answers</strong></p><p>&ndash;76.2 uses 1 mol of electrons instead of 2</p><p>+152 uses &ndash;0.79 V instead of +0.79 V (or emits to use the &ndash;ve sign in the equation)</p><p>+76.2 uses 1 mol of electrons instead of 2 <strong>and</strong> uses &ndash;0.79 V instead of +0.79 V (or emits to use the &ndash;ve sign in the equation)</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>Which half-equations occur at the negative and positive electrodes respectively when aqueous copper chloride solution is electrolysed using copper electrodes?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>negative: Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> &rarr; Cu<sub>(s)</sub> positive: Cu<sub>(s)</sub> &rarr; Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> </span></label></p><p><label class="radio"> <input type="radio"> <span>negative: Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> &rarr; Cu<sub>(s)</sub> positive: Cl<sup>&ndash;</sup><sub>(aq)</sub> &rarr; &frac12;Cl<sub>2(g)</sub> + e<sup>&ndash;</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>negative: Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> &rarr; Cu<sub>(s)</sub> positive: H<sub>2</sub>O<sub>(l)</sub> &rarr; &frac12;O<sub>2</sub> + 2H<sup>+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>negative H<sub>2</sub>O<sub>(l)</sub> + e<sup>&ndash;</sup> &rarr; &frac12;H<sub>2(g)</sub> + OH<sup>&ndash;</sup><sub>(aq)</sub> positive: H<sub>2</sub>O<sub>(l)</sub> &rarr; &frac12;O<sub>2</sub> + 2H<sup>+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> </span></label> </p></div><div class="q-explanation"><p>The species discharged at the negative electrode, gaining e<sup>&ndash;</sup> will be lower (less &ndash;ve) in the electrochemical series; copper is lower than water so copper ions are discharged.</p><p>Normally the species discharged at the positive electrode, losing e<sup>&ndash;</sup>, will be higher (less +ve) in the electrochemical series. Note! Exception: Chloride ions (&frac12;Cl<sub>2(g)</sub> + e<sup>&ndash;</sup> ⇌ Cl<sup>&ndash;</sup><sub>(aq) </sub>E<sup><s>o</s></sup> = +1.36 V) are preferentially discharged over water (&frac12;O<sub>2(g)</sub> + 2H<sup>+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ H<sub>2</sub>O<sub>(l)</sub> E<sup><s>o</s></sup> = +1.23 V) at the positive electrode despite their E<sup><s>o</s></sup> values (this is a kinetic effect that needs remembering).</p><p>However, if a copper electrode is used instead of an inert electrode, copper from the electrode will lose electrons preferentially over chloride ions or water (Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ Cu<sub>(s)</sub> E<sup><s>o</s></sup> = +0.34 V) because its E<sup><s>o</s></sup> is higher in the electrochemical series.</p><p>Thus <strong>negative: Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> &rarr; Cu<sub>(s)</sub> positive: Cu<sub>(s)</sub> &rarr; Cu<sup>2+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> </strong>is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is produced at the negative and positive electrodes respectively when aqueous copper(II) chloride solution is electrolysed using inert electrodes?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>hydrogen and oxygen</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>copper and chlorine</span></label> </p><p><label class="radio"> <input type="radio"> <span>copper and oxygen</span></label> </p><p><label class="radio"> <input type="radio"> <span>hydrogen and chlorine</span></label> </p></div><div class="q-explanation"><p>The species discharged at the negative electrode, gaining e<sup>&ndash;</sup> will be lower (less &ndash;ve) in the electrochemical series; copper is lower than water so copper ions are discharged.</p><p>The species discharged at the positive electrode, losing e<sup>&ndash;</sup>, will be higher (less +ve) in the electrochemical series. Note! Exception: Chloride ions (&frac12;Cl<sub>2(g)</sub> + e<sup>&ndash;</sup> ⇌ Cl<sup>&ndash;</sup><sub>(aq) </sub>E<sup><s>o</s></sup> = +1.36 V) are preferentially discharged over water (&frac12;O<sub>2(g)</sub> + 2H<sup>+</sup><sub>(aq)</sub> + 2e<sup>&ndash;</sup> ⇌ H<sub>2</sub>O<sub>(l)</sub> E<sup><s>o</s></sup> = +1.23 V) at the positive electrode despite their E<sup><s>o</s></sup> values (this is a kinetic effect that needs remembering).</p><p>Thus <strong>copper and chlorine</strong> is the correct answer.</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Calculator question: </em>What mass of zinc (in g to 3 sf) is deposited at the cathode if zinc sulfate solution is electrolysed for 30.0 minutes at a current of 5.00A?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>3.05</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.0508</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.102</span></label> </p><p><label class="radio"> <input type="radio"> <span>6.10</span></label> </p></div><div class="q-explanation"><p>Charge = current &times; time = 5.0 &times; (30.0 &times; 60) = 9000C</p><p>Moles of electrons passed through circuit = charge/Faraday = 9000/96500 = 0.0932642</p><p>Half equation: Zn<sup>2+</sup><sub>(aq) </sub>+ 2e<sup>&ndash;</sup> &rarr; Zn(s) so mole ratio of electrons to zinc atoms is 2:1,</p><p>Therefore mass of zinc = moles &times; Ar = 0.0932642/2 &times; 65.38 = 3.0488083 = <strong>3.05g</strong> (3sf)</p><p>Therefore <strong>3.05</strong> is the correct answer.</p><p><strong>Incorrect answers</strong></p><p>6.10 uses a 1:1 mole ratio for electrons to atoms instead of 2:1</p><p>0.0508 uses 150C of charge (forgetting to multiply minutes by 60 to get seconds)</p><p>0.102 uses 150C of charge (forgetting to multiply minutes by 60 to get seconds) <strong>and</strong> uses a 1:1 mole ratio for electrons to atoms instead of 2:1</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p><em>Calculator question: </em>Two cells are electrolysed in series. What mass of calcium (in g to 3 sf) is deposited at the cathode in &lsquo;cell A&rsquo;, if 1.62g aluminium is deposited at the cathode in &lsquo;cell B&rsquo;?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2.41</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>3.61</span></label> </p><p><label class="radio"> <input type="radio"> <span>4.81</span></label> </p><p><label class="radio"> <input type="radio"> <span>1.60</span></label> </p></div><div class="q-explanation"><p>Moles of aluminium = mass/Ar = 1.62/26.98 = 0.0600444 mol</p><p>Half equations: Al<sup>3+</sup><sub>(aq)</sub> + <u>3e</u><sup>&ndash;</sup> &rarr; Al<sub>(s) </sub> and Ca<sup>2+</sup><sub>(aq)</sub> + <u>2e</u><sup>&ndash;</sup> &rarr; Ca<sub>(s)</sub> so ratio of<strong> moles of electrons</strong> for aluminium to <strong>moles of electrons</strong> for calcium is 3:2.</p><p>This means that the ratio of moles of aluminium:calcium will be 2:3 (since the same number of electrons will pass through the circuit, but that will give more calcium atoms as only two electrons are needed for each atom, rather than 3 for aluminium).</p><p>Moles of calcium = 0.0600444 &times; 3/2 = 0.0900666</p><p>Mass of calcium = moles &times; Ar = 0.0900666 &times; 40.08 = 3.6098693 = 3.61g (3sf)</p><p>Therefore <strong>3.61</strong> is the correct answer.</p><p><strong>Incorrect answers</strong></p><p>1.60 uses a 3:2 mole ratio for moles of aluminium:calcium instead of 2:3</p><p>2.41 uses a 1:1 mole ratio for moles of aluminium:calcium instead of 2:3</p><p>4.81 uses a 1:2 mole ratio for moles of aluminium:calcium instead of 2:3</p></div><div class="actions"><span class="score" data-score="0"></span><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </div> </div> </div> </div> <div class="panel-footer"> <div>&nbsp;</div> </div> </div> <div class="panel panel-has-colored-body panel-default"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Exam-style questions</p> </div> </div> <div class="panel-body"> <div> <h4>Paper 1</h4> <h5>Core (SL&amp;HL):&nbsp;&nbsp;&nbsp;<a href="../2724/redox-core-sl-and-hl-paper-1-questions.html" title="Redox core (SL and HL) paper 1 questions">Redox core (SL and HL) paper 1 questions</a></h5> <h5>AHL (HL only):&nbsp;&nbsp;&nbsp;&nbsp;<a href="../2731/redox-ahl-hl-only-paper-1-questions.html" title="Redox AHL (HL only) paper 1 questions">Redox AHL (HL only) paper 1 questions</a></h5> <h4>Paper 2</h4> <h5>Core (SL&amp;HL):&nbsp;&nbsp;&nbsp;<a href="../2728/redox-core-sl-hl-paper-2-questions.html" title="Redox core (SL &amp; 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