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pull-right" data-pid="905" 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">Oxidation and reduction</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 45 minutes"><i class="fa fa-clock-o"></i> 45'</span> </ol> <article id="main-article"> <p><img alt="" src="../../images/redox-processes/blastfurnace-1.jpg" style="width: 160px; height: 131px; float: left;">It is important to understand that redox is about electron counting. If you understand our oxidation state model, then the rest of the Redox topic will be more accessible and less confusing. We are always asking ourselves <em>'Which atoms have the valance electrons?'</em>. In ionic compounds that's easy to answer (see<strong> Ionic bonding</strong> 4.1), since in our model electrons transfer. In covalent compounds we assign the electrons in bonds to the atoms with the highest electronegativity. This is the basis of our oxidation state model - keep that in mind; despite all the stuff to learn, the principle of oxidation and reduction is a simple one.</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>Redox, <strong>Oxidation*, Reduction*, Oxidising agent*, Reducing agent*, </strong>Oxidation state, Oxidation number, Half-equation, Activity series, The Winkler method, Biological oxygen demand</p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="1" data-mode="Normal" data-topics="629" data-subject-id="7" data-n-flashcards="11" style="text-align:center">Show flashcards</a></div><hr> <div class="panel panel-has-colored-body panel-has-border panel-turquoise"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Redox unwrapped</p> </div> </div> <div class="panel-body"> <div>Understanding our model of oxidation states and how it pulls together definitions of oxidation and reduction.</div> <div> <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/446183341"></iframe></div> <p> </p> </div> </div> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div> <p> The revision cards contain all of the essential content:</p> <div id="carousel-202" class="dynamic-gallery carousel slide" data-id="202"><div class="carousel-inner" role="listbox"><div class="item active"><a class="fancy" href="../../../std-galleries/7-202/screenshot-2020-08-09-at-110722.png" data-fancybox="gallery-202" title="" data-caption=""><img alt="" src="../../../std-galleries/7-202/screenshot-2020-08-09-at-110722.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-202/screenshot-2020-08-12-at-215116.png" data-fancybox="gallery-202" title="" data-caption=""><img alt="" src="../../../std-galleries/7-202/screenshot-2020-08-12-at-215116.png"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/7-202/screenshot-2020-08-09-at-102610.png" data-fancybox="gallery-202" title="" data-caption=""><img alt="" src="../../../std-galleries/7-202/screenshot-2020-08-09-at-102610.png"></a></div><div class="item "><a class="fancy" 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carousel-control" href="#carousel-202" role="button" data-slide="prev"><i class="fa fa-fw fa-chevron-left"></i></a><a class="right carousel-control" href="#carousel-202" role="button" data-slide="next"><i class="fa fa-fw fa-chevron-right"></i></a></div><ol class="std-carousel-indicators"><li data-index="0"><img title="Click to view" src="../../../std-galleries/7-202/screenshot-2020-08-09-at-110722-thumb128.jpg"><li><li data-index="1"><img title="Click to view" src="../../../std-galleries/7-202/screenshot-2020-08-12-at-215116-thumb128.jpg"><li><li data-index="2"><img title="Click to view" src="../../../std-galleries/7-202/screenshot-2020-08-09-at-102610-thumb128.jpg"><li><li data-index="3"><img title="Click to view" src="../../../std-galleries/7-202/screenshot-2020-08-09-at-102625-thumb128.jpg"><li><li data-index="4"><img title="Click to view" src="../../../std-galleries/7-202/screenshot-2020-08-09-at-102653-thumb128.jpg"><li><li data-index="5"><img title="Click to view" 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</p><p><label class="radio"> <input class="c" 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 only</span></label> </p></div><div class="q-explanation"><p>'1 and 2 only' is the correct answer.</p><p>Oxidation can be defined in terms of an element gaining oxygen or losing hydrogen (both useful in organic chemistry), or may be defined as <strong>loss of electrons </strong>(OILRIG) or an increase in oxidation state.</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>Which of the following describes reduction?</p><p><strong>1: </strong>The loss of oxygen or gain of hydrogen.</p><p><strong>2: </strong>A decrease in oxidation state.</p><p><strong>3: </strong>The gain of electrons.</p></div><div class="q-answer"><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 3 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 and 2 only</span></label> </p></div><div class="q-explanation"><p>Reduction can be defined in all of these ways. In terms of an element losing oxygen or gaining hydrogen (both useful in organic chemistry), or gaining electrons<strong> </strong>(OILRIG) or a decrease in oxidation state.</p><p>Thus 1, 2 and 3 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>Which of the following reactions could theoretically occur as written, given sufficient activation energy?</p><p><strong>1: </strong>Mg + Zn(NO<sub>3</sub>)<sub>2</sub> → Mg(NO<sub>3</sub>)<sub>2</sub> + Zn</p><p><strong>2: </strong>Zn + MgSO<sub>4</sub> → ZnSO<sub>4</sub> + Mg<strong> </strong></p><p><strong>3: </strong>CuSO<sub>4</sub> + Zn → ZnSO<sub>4</sub> + Cu</p></div><div class="q-answer"><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 2 only</span></label> </p><p><label class="radio"> <input type="radio"> <span>3 only</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1 and 3 only</span></label> </p></div><div class="q-explanation">These redox reactions are simple replacement (displacement) reactions. An activity series is given in the data book. 1 and 3 could occur as magnesium is more reactive than zinc, and zinc is more reactive than copper (more reactive meaning more likely to lose electrons). But zinc is not more reactive than magnesium, so reaction 2 could not occur (see data book).</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>What is the oxidation state of chromium in potassium dichromate K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>+6</span></label> </p><p><label class="radio"> <input type="radio"> <span>12+</span></label> </p><p><label class="radio"> <input type="radio"> <span>+12</span></label> </p><p><label class="radio"> <input type="radio"> <span>6+</span></label> </p></div><div class="q-explanation"><p>+6 is the correct answer, since each oxygen atom has an oxidation state of −2 and each potassium +1.</p><p>(7 × −2) + (2 × +1) = −12 and the compound is neutral overall (0), so the two chromium atoms carry +12, thus +6 for each atom. The oxidation state of chromium is +6.</p><p>Note that 6+ is a <strong>charge</strong>, oxidation states must be shown with the + or − sign before the number. When oxidation states are assigned, it is as if everything is ionic, but this is not the case.</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>What is the oxidation state of chlorine in the chlorate ion ClO<sub>3</sub><sup>−</sup>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>+1</span></label> </p><p><label class="radio"> <input type="radio"> <span>−1</span></label> </p><p><label class="radio"> <input type="radio"> <span>+3</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>+5</span></label> </p></div><div class="q-explanation"><p>+5 is the correct answer, since oxygen is the most electronegative element and will therefore carry the negative oxidation state; each oxygen atom has an oxidation state of −2.</p><p>(3 × −2) = −6 and the compound has a 1− <strong>charge</strong> (notice for the charge the number comes first) which means the ion has an overall oxidation state of −1, so the chlorine atom has an oxidation state of +5 (−6 + +5 = −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">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the <strong>oxidation number </strong>of iron in Fe<sub>2</sub>O<sub>3</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>+2</span></label> </p><p><label class="radio"> <input type="radio"> <span>3+</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>III</span></label> </p><p><label class="radio"> <input type="radio"> <span>+3</span></label> </p></div><div class="q-explanation"><p>'III' is the correct answer; <strong>oxidation numbers</strong> are represented by Roman numerals (e.g. iron (III) oxide or copper (II) oxide) and <strong>oxidation states</strong> by numbers with the charge in front.</p><p>In Fe<sub>2</sub>O<sub>3</sub> each oxygen has an oxidation state of −2; 3 × −2 = −6 and the compound is neutral overall (0), so the two irons carry +6, thus +3 for each. The oxidation state of iron is +3. And this compound is ionic, so the<strong> charge</strong> on the iron ion is 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="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>What is the name of the compound MnO<sub>2</sub>?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>Manganese (IV) oxide</span></label> </p><p><label class="radio"> <input type="radio"> <span>Manganese (VII) oxide</span></label> </p><p><label class="radio"> <input type="radio"> <span>Manganese oxide (II)</span></label> </p><p><label class="radio"> <input type="radio"> Manganese (II) oxide<span></span></label> </p></div><div class="q-explanation"><p>'Manganese (IV) oxide' is the correct answer.</p><p>In MnO<sub>2</sub> each oxygen atom has an oxidation state of −2; 2 × −2 = −4 and the compound is neutral overall (0), so the manganese atom has an oxidation state of +4, this is represented by the oxidation number of IV in the name.</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>Iron (II) ions react with hydrogen peroxide in acidic solution according to the following equation:</p><p>2Fe<sup>2+</sup> + H<sub>2</sub>O<sub>2</sub> + 2H<sup>+</sup> → 2Fe<sup>3+</sup> + 2H<sub>2</sub>O</p><p>Which chemical species is the reducing agent in this reaction?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>H<sup>+</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>H<sub>2</sub>O<sub>2</sub></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>Fe<sup>2+</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>Fe<sup>3+</sup></span></label> </p></div><div class="q-explanation"><p>A reducing agent reduces another element and therefore is itself oxidised (OILRIG: if it gives electrons away to another element, it has lost electrons).</p><p>Oxidation state changes are needed:</p><p>Iron goes from +2 to +3 (Fe<sup>2+</sup> to Fe<sup>3+</sup>).</p><p>Oxygen goes from −1 (always in a peroxide) to −2 (H<sub>2</sub>O<sub>2</sub> to H<sub>2</sub>O).</p><p>Hydrogen is +1 throughout.</p><p>Thus iron is oxidised and oxygen is reduced. The Fe<sup>2+</sup> ions are the reducing agent; 'giving an electron' to oxygen.</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>Balance the following redox equation to give the lowest whole number ratio:</p><p>____OH<sup>−</sup> + ____SO<sub>3</sub><sup>2−</sup> + ____Fe<sup>3+</sup> → ____SO<sub>4</sub><sup>2−</sup> + ____H<sub>2</sub>O + ____Fe<sup>2+</sup></p><p>How many moles of iron (III) ions are required?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>0</span></label> </p><p><label class="radio"> <input type="radio"> <span>4</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>2</span></label> </p><p><label class="radio"> <input type="radio"> <span>1</span></label> </p></div><div class="q-explanation"><p>Always balance the electrons first by calculating changes in oxidation states:</p><p>Oxygen remains as −2 throughout.</p><p>Hydrogen remains as +1 throughout.</p><p>Sulfur changes from +4 to +6 (loss of 2 electrons).</p><p>Iron changes from +3 to +2 (gain of 1 electron).</p><p>To balance electrons lost and gained, <strong>two</strong> iron ions are needed:</p><p>____OH<sup>−</sup> + ____SO<sub>3</sub><sup>2−</sup> + <strong>2</strong>Fe<sup>3+</sup> → ____SO<sub>4</sub><sup>2−</sup> + ____H<sub>2</sub>O + <strong>2</strong>Fe<sup>2+</sup></p><p>Now atoms of other elements can be balanced:</p><p><strong>2</strong>OH<sup>−</sup> + SO<sub>3</sub><sup>2−</sup> + <strong>2</strong>Fe<sup>3+</sup> → SO<sub>4</sub><sup>2−</sup> + H<sub>2</sub>O + <strong>2</strong>Fe<sup>2+</sup></p><p>So the correct answer is '2'.</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>Balance the following redox equation to give the lowest whole number ratio:</p><p>____H<sup>+</sup> + ____BiO<sub>3</sub><sup>−</sup> + ____Mn<sup>2+</sup> → ____Bi<sup>3+</sup> + ____H<sub>2</sub>O + ____MnO<sub>4</sub><sup>−</sup></p><p>How many moles of H<sup>+</sup> ions are required?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>12</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>14</span></label> </p><p><label class="radio"> <input type="radio"> <span>4</span></label> </p><p><label class="radio"> <input type="radio"> <span>2</span></label> </p></div><div class="q-explanation"><p>Always balance the electrons first by calculating changes in oxidation states:</p><p>Oxygen remains as −2 throughout.</p><p>Hydrogen remains as +1 throughout.</p><p>Bismuth (Bi) changes from +5 to +3 (gain of 2 electrons).</p><p>Manganese (Mn) changes from +2 to +7 (loss of 5 electrons).</p><p>To balance electrons lost and gained (10 electrons), <strong>five</strong> 'bismuths' are needed and <strong>two </strong>'manganeses':</p><p>____H<sup>+</sup> + <strong>5</strong>BiO<sub>3</sub><sup>−</sup> + <strong>2</strong>Mn<sup>2+</sup> → <strong>5</strong>Bi<sup>3+</sup> + ____H<sub>2</sub>O + <strong>2</strong>MnO<sub>4</sub><sup>−</sup></p><p>Now atoms of other elements can be balanced:</p><p>Oxygens:</p><p>____H<sup>+</sup> + 5BiO<sub>3</sub><sup>−</sup> + 2Mn<sup>2+</sup> → 5Bi<sup>3+</sup> + <strong>7</strong>H<sub>2</sub>O + 2MnO<sub>4</sub><sup>−</sup></p><p>Then hydrogens:</p><p><strong>14</strong>H<sup>+</sup> + 5BiO<sub>3</sub><sup>−</sup> + 2Mn<sup>2+</sup> → 5Bi<sup>3+</sup> + 7H<sub>2</sub>O + 2MnO<sub>4</sub><sup>−</sup></p><p>So the correct answer is '14'.</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">11</div><div class="exercise shadow-bottom"><div class="q-question"><p>The Winkler method is carried out to find the dissolved oxygen content of a sample of cold river water.</p><p>250cm<sup>3</sup> of the water is taken and treated with manganese (II) chloride, sodium iodide, sodium hydroxide and finally sulphuric acid, to yield iodine. The iodine is titrated against thiosulphate ions.</p><p>On average 4.20 × 10<sup>−5</sup> moles of thiosulphate ions are used in the redox titration for a 25.0 cm<sup>3</sup> aliquot of the 'water solution'.</p><p>How many moles of dissolved oxygen are there in a 25.0 cm<sup>3</sup> sample of the warm water?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2.10 × 10<sup>−5</sup></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1.05 × 10<sup>−5</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>8.40 × 10<sup>−5</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>4.20 × 10<sup>−5</sup></span></label> </p></div><div class="q-explanation"><p>The moles of dissolved oxygen in the aliquot is always <strong>one quarter</strong> of the moles of thiosulphate used in the titration:</p><p>4.20 × 10<sup>−5</sup> / 4 = <strong>1.05 × 10</strong><sup>−5</sup><strong> moles</strong></p><p>This, incidently, equates to 13.4 mg dm<sup>−3</sup> which is the same as<strong> parts per million </strong>(in fact, mg per<sup> </sup>1000g (1 dm<sup>3</sup>) of solution). A value of 13.4 ppm suggests that the river water is not likely to be polluted.</p><p>Working backwards through the Winkler method:</p><p>2S<sub>2</sub>O<sub>3</sub><sup>2-</sup>(aq) + I<sub>2</sub> <img height="9" class="gifffer" data-gifffer="http://ibchem.com/IB16/img/arrow.gif" width="36"> S<sub>4</sub>O<sub>6</sub><sup>2-</sup>(aq) + 2I<sup>-</sup>(aq)</p><p>2 moles of thiosulphate reacts with <span style="font-size:12.0pt;font-family:Cambria;
mso-fareast-font-family:">1</span> mole of iodine (I<sub>2</sub>).</p><p>2Mn<sup>3+</sup>(aq) + 2I<sup>-</sup>(aq) <img height="9" class="gifffer" data-gifffer="http://ibchem.com/IB16/img/arrow.gif" width="36"> I<sub>2</sub> + 2Mn<sup>2+</sup>(aq)</p><p>1 mole of iodine is liberated by 2 moles of Mn<sup>3+ </sup>ions.</p><p>Mn(OH)<sub>3</sub>(s) + 3H<sup>+</sup>(aq) <img height="9" class="gifffer" data-gifffer="http://ibchem.com/IB16/img/arrow.gif" width="36"> Mn<sup>3+</sup>(aq) + 3H<sub>2</sub>O</p><p>2 moles of Mn<sup>3+</sup> ions are produced by 2 moles of manganese (III) hydroxide (1:1 ratio).</p><p>4Mn(OH)<sub>2</sub> + O<sub>2</sub> + 2H<sub>2</sub>O <img height="9" class="gifffer" data-gifffer="http://ibchem.com/IB16/img/arrow.gif" width="36"> 4Mn(OH)<sub>3</sub></p><p>2 moles of manganese (III) hydroxide is produced from ½ mole (ratio is 4:1) of dissolved oxygen.</p><p>So overall 4 moles of thiosulphate ions equates to 1 mole of dissolved oxygen.</p></div><div 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