{"id":5961,"date":"2020-10-13T12:02:29","date_gmt":"2020-10-13T06:32:29","guid":{"rendered":"http:\/\/astan.lk\/al_virtualclassroom\/?p=5961"},"modified":"2020-10-13T12:02:33","modified_gmt":"2020-10-13T06:32:33","slug":"calculations","status":"publish","type":"post","link":"https:\/\/astan.lk\/al_virtualclassroom\/calculations\/","title":{"rendered":"Calculations"},"content":{"rendered":"

A chemical equation should be balanced with respect to mass and charges of ions in order to get quantitative information from it.<\/p>\n

balancing simple nuclear equation<\/p>\n

\"images<\/a><\/p>\n

 <\/p>\n

Inspection method<\/strong><\/span><\/p>\n

Balancing an equation by balancing the number of atoms of each kind in the reactants and products is called inspection balancing.<\/p>\n

Balancing redox equations by checking the changes in oxidation number<\/strong><\/span><\/p>\n

One way to balance redox reactions is by keeping track of the electron transfer using the oxidation numbers of each of the atoms. For the oxidation-number-change method<\/strong>, start with the unbalanced skeleton equation. The example below is for the reaction of iron(III) oxide with carbon monoxide. This reaction is one that takes place in a blast furnace during the processing of iron ore into metallic iron<\/p>\n

Fe<\/span>2<\/span><\/span>O<\/span>3<\/span><\/span>(<\/span>s<\/span>)<\/span>+<\/span>CO<\/span>(<\/span>g<\/span>)<\/span>\u2192<\/span>Fe<\/span>(<\/span>s<\/span>)<\/span>+<\/span>CO<\/span>2<\/span><\/span>(<\/span>g<\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

Step 1:<\/strong> Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom.<\/em><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0+3 \u00a0 -2 \u00a0 \u00a0 \u00a0 \u00a0+2 -2 \u00a0 \u00a0 \u00a00 \u00a0 \u00a0 \u00a0 \u00a0+4 -2 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Fe<\/span>2<\/span><\/span><\/span>O<\/span>3<\/span><\/span><\/span>(<\/span>s<\/span>)<\/span>+<\/span>C\u00a0<\/span><\/span>O<\/span><\/span>(<\/span>g<\/span>)<\/span>\u2192<\/span>Fe<\/span><\/span>(<\/span>s<\/span>)<\/span>+<\/span>C<\/span><\/span>O<\/span>2<\/span><\/span><\/span>(<\/span>g<\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/em><\/p>\n

Step 2:<\/strong> Identify the atoms that are oxidized and those that are reduced<\/em>. In the above equation, the carbon atom is being oxidized since its oxidation increases from +2 to +4. The iron atom is being reduced since its oxidation number decreases from +3 to 0.<\/p>\n

Step 3:<\/strong> Use a line to connect the atoms that are undergoing a change in oxidation number.<\/em> On the line, write the oxidation-number change.<\/p>\n

\"Walkthrough<\/span><\/p>\n

The carbon atom\u2019s oxidation number increases by 2, while the iron atom\u2019s oxidation number decreases by 3. As written, the number of electrons lost does not equal the number of electrons gained. In a balanced redox equation, these must be equal. So, the increase in oxidation number of one atom must be made equal to the decrease in oxidation number of the other.<\/p>\n

Step 4:<\/strong> Use coefficients to make the total increase in oxidation number equal to the total decrease in oxidation number<\/em>. In this case, the least common multiple of 2 and 3 is 6. So the oxidation-number increase should be multiplied by 3, while the oxidation-number decrease should be multiplied by 2. The coefficient is also applied to the formulas in the equation. So a 3 is placed in front of the CO and in front of the CO2<\/sub>. A 2 is placed in front of the Fe on the right side of the equation. The Fe2<\/sub>O3<\/sub> does not require a coefficient because the subscript of 2 after the Fe indicates that there are already two iron atoms.<\/p>\n

\"Walkthrough<\/span><\/p>\n

Step 5:<\/strong> Check the balancing for both atoms and charge.<\/em> Occasionally, a coefficient may need to be placed in front of a molecular formula that was not involved in the redox process. In the current example, the equation is now balanced.<\/p>\n

Fe<\/span>2<\/span><\/span>O<\/span>3<\/span><\/span>(<\/span>s<\/span>)<\/span>+<\/span>3<\/span>CO<\/span>(<\/span>g<\/span>)<\/span>\u2192<\/span>2<\/span>Fe<\/span>(<\/span>s<\/span>)<\/span>+<\/span>3<\/span>CO<\/span>2<\/span><\/span>(<\/span>g<\/span>)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Balancing equations using oxidation \/ reduction half reactions<\/span><\/strong><\/p>\n

Balancing redox reactions is slightly more complex than balancing standard reactions, but still follows a relatively simple set of rules. One major difference is the necessity to know the half-reactions of the involved reactants; a half-reaction table is very useful for this. \u00a0Half-reactions are often useful in that two half reactions can be added to get a total net equation. Although the half-reactions must be known to complete a redox reaction, it is often possible to figure them out without having to use a half-reaction table. This is demonstrated in the acidic and basic solution examples. Besides the general rules for neutral conditions, additional rules must be applied for aqueous reactions in acidic or basic conditions.<\/p>\n

The method used to balance redox reactions is called the Half Equation Method<\/strong>. In this method, the equation is separated into two half-equations; one for oxidation and one for reduction.<\/p>\n

Each equation is balanced by adjusting coefficients and adding H2<\/sub>O, H+<\/sup>, and e–<\/sup> in this order:<\/p>\n

    \n
  1. Balance elements in the equation other than O and H.<\/li>\n
  2. Balance the oxygen atoms by adding the appropriate number of water (H2<\/sub>O) molecules to the opposite side of the equation.<\/li>\n
  3. Balance the hydrogen atoms\u00a0(including those added in step 2 to balance the oxygen atom) by adding H+<\/sup> ions to the opposite\u00a0side of the equation.<\/li>\n
  4. Add up the charges on each side. Make them equal by adding enough electrons (e–<\/sup>) to the more positive side. (Rule of thumb: e–<\/sup> and H+<\/sup> are almost always on the same side.)<\/li>\n
  5. The e–<\/sup> on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers (the lowest common multiple) to be made the same.<\/li>\n
  6. The half-equations are added together, canceling out the electrons to form one balanced equation. Common terms should also be canceled out.<\/li>\n<\/ol>\n
      \n
    • (If the equation is being balanced in a basic solution, through the addition of one more step, the appropriate number of OH–<\/sup> must be added to turn the remaining H+<\/sup> into water molecules.)<\/li>\n
    • The equation can now be checked to make sure that it is balanced.<\/li>\n<\/ul>\n

       <\/p>\n","protected":false},"excerpt":{"rendered":"

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