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Working Out Formulae & Balancing Equations (including electrolysis)

Working Out Formulae & Balancing Equations (including electrolysis). Valency. Writing Equations. Using Valency. Balancing Equations. Electrolysis half equations. ALWAYS BRINGS YOU BACK HERE. END. VALENCY tells you how many BONDS an atom can form. You work the valency out like this….

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Working Out Formulae & Balancing Equations (including electrolysis)

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  1. Working Out Formulae & Balancing Equations (including electrolysis) Valency Writing Equations Using Valency Balancing Equations Electrolysis half equations ALWAYS BRINGS YOU BACK HERE END

  2. VALENCY tells you how many BONDS an atom can form You work the valency out like this… Work out how many OUTER ELECTRONS the atom has (That’s it’s GROUP NUMBER in the Periodic Table) If it has 4 or less then THAT’S THE VALENCY If it has 5 or more then VALENCY = 8 – OUTER ELECTRONS

  3. Here’s another method that works for ions… Sodium forms Na+ ion It has valency 1 For IONIC BONDS, the valency is just the size of the charge on the ion Calcium forms Ca2+ ion It has valency 2 Oxygen forms O2-ion It has valency 2 Chlorine forms Cl-ion It has valency 1 Some transition metals can form more than one ion, so they have more than one valency. Iron forms Fe2+ and Fe3+ So it can have valency 2 or 3 You can use the same idea for ions made of more than one atom Ammonium ion NH4+ has valency 1 Nitrate ion NO3- has valency 1 Sulphate ion SO42- has valency 2

  4. USING VALENCIES TO WORK OUT FORMULAE • Write down the two elements (or ions) in the compound • Write down the valency of each as a “small number” next to the other one. Put brackets round any complicated ions, like sulphate, nitrate etc. • Cancel if necessary, and remove any 1s SodiumsulphateNaSO4 Valency of sodium: 1 Valency of sulphate: 2 So Na2(SO4)1 Remove 1s: Na2SO4 Calciumoxide: CaO Valency of calcium: 2 Valency of oxygen: 2 So Ca2O2 Cancel: CaO Iron(II) chlorideFe Cl Valency of iron : 2 Valency of chloride : 1 So Fe1Cl2 Remove 1s: Fe Cl2 Magnesium nitrate Mg NO3 Valency of magnesium : 2 Valency of nitrate : 1 So Mg1(NO3)2 Remove 1s: Mg (NO3)2 DO NOT “MULTIPLY OUT THE BRACKETS” ! Write (NO3)2 not N2O6

  5. WRITING EQUATIONS Reactants Products carbon dioxide calcium chloride calcium carbonate hydrochloric acid + + water + Step 1: Write a WORD EQUATION What you make in the reaction The chemicals you start with Step 2: Write down the FORMULA of each of the chemicals in your word equation Step 3: Put in the STATE SYMBOLS (s) for solid, (l) for liquid, (aq) for solution, (g) for gas Eg: reaction of calcium carbonate with hydrochloric acid: The PRODUCTS CaCl2 + H2O + CO2 CaCO3 + HCl The REACTANTS (g) (aq) (aq) (s) (l) Make sure you get the formulae right! No marks for balancing if the formulae are wrong! Calcium carbonate is an insoluble solid Hydrochloric acid, like other acids, is always a solution Now on to the harder bit – balancing the equation! Calcium chloride is soluble, so as there is water present, it is a solution Water is a liquid (not a solution!) Carbon dioxide is a gas It forms in bubbles

  6. Balancing Equations Atom Number on Left Number on Right WHAT DOES IT MEAN? Balancing an equation means making sure the numbers of each type of atom are the same on each side Let’s look at this balanced equation: 2Na + 2H2O 2NaOH + H2 1 in each NaOH 2NaOH. So 2 Na Have 2Na, so 2 2 in each water 2 waters 2  2 = 4 H 2 from 2NaOH + 2 from H2 makes 4 1 in each water 2 waters. So 2 1 in each NaOH 2NaOH. So 2 O They all match. So it’s balanced!

  7. How to balance an equation Na is only in one chemical each side Step 1: Write down the unbalanced formula equation. Step 2: Work out how many of each atom there are on each side. (in your head, if it’s easy) Step 3: Look for any atoms where there aren’t the same number on each side Step 4: Choose the “unbalanced atom” that’s in the smallest number of different formulae. Step 5: Balance it by putting a number IN FRONT of one of the formulae (don’t change the actual formula!) Step 6: Recalculate numbers of atoms – and repeat if needed! NaOH + H2SO4 Na2SO4 + H2O 2 2 Left Right Na 1 2 O 5 5 H 3 2 S 1 1 2 There are different numbers of O and H 6 6 There are different numbers of Na and H H is in fewer different formulae. 4 4 We can balance H by putting 2 in front of H2O We can balance them by putting 2 in front of NaOH THEY ALL MATCH! IT’S BALANCED! Now recalculate the numbers of atoms… Now recalculate the numbers of atoms…

  8. Other examples… 2 Let’s balance the hydrogens: We can do this by putting 2 in front of HCl   2 2 balanced balanced Let’s balance the chlorines : We can do this by putting 2 in front of KCl Unfortunately, this has also unbalanced the potassium. However putting 2 in front of KI balances both K and I Left Right Ca 1 1 C 1 1 O 3 3 H 1 2 Cl 1 2 CaCO3 + HCl  CaCl2 + H2O + CO2 2 2 Eureka ! That has automatically balanced the chlorines too. Left Right Cl 2 1 K 1 1 I 1 2 Cl2 + KI  KCl + I2 2 2 2 2

  9. Another example… 2 3 Al2O3 + HCl  AlCl3 + H2O 6 Left Right Al 2 1 O 3 1 H 1 2 Cl 1 3 2 Now let’s balance the oxygens: We can do this by putting 3 in front of H2O Now let’s balance the hydrogens: We can do this by putting 6 in front of HCl 3 None of the atoms are balanced! They all occur in just two chemicals Choose one to balance… 6 6 6 6 Recalculate… Recalculate… We can balance Al by putting 2 in front of AlCl3 THEY ALL MATCH! IT’S BALANCED! Recalculate…

  10. An awkward one! 2 2 Al + Cl2 AlCl3 3 Left Right Al 1 1 Cl 2 3 2 2 Now we must balance the aluminiums We can do this by putting 2 in front of Al Chlorines aren’t balanced. 6 6 But how can we do the balancing? We haven’t got “nice” numbers! Recalculate… This is like finding the “lowest common denominator” in fractions. We have two chlorines on one side, and three on the other. We find the smallest number two and three go into – that’s six. So we need to aim for six chlorines on each side THEY ALL MATCH! IT’S BALANCED! If you like maths, you could try balancing ones like this using fractions instead. You’d need to use 1½ To do that, we put 3 in front of Cl2 and 2 in front of AlCl3 Recalculate…

  11. When electrons are gained by a positive ion, the name of the chemical change is REDUCTION. REDUCTION IS THE GAIN OF ELECTRONS. THE COPPER ION HAS BEEN REDUCED Cu The electrode is negative because it has too many electrons Cu2+ Cu2+ Cu2+ Cu2+ e- go to ion 2e- + Cu2+ Cu e- go to ion. TWO ELECTRONS FROM THE CATHODE A NEUTRAL ATOM OF THE ELEMENT COPPER. This makes copper the element, which covers the electrode. ARE ADDED TO THE COPPER ION ELECTROLYSIS Here are Cu2+ ions moving to the negative electrode. Positive ions in the solution are attracted to negative electrode – opposite charges attract. As they get close, the ions gain electrons from the electrode and the Cu2+ is neutralised.

  12. e- go to cell Cl- When electrons are lost by a negative ion, the name of the chemical change is OXIDATION. OXIDATION IS THE LOSS OF ELECTRONS THE CHLORIDE ION HAS BEEN OXIDISED the ion the ion Cl- loses an e- loses an e- This electrode is positive because some electrons have been removed by the cell. Cl- Cl2 Cl- Cl- TWO CHLORIDE IONS, EACH WITH AN EXTRA ELECTRON THE TWO ELECTRONS LOST BY THE IONS GO TO THE ELECTRODE A NEUTRAL CHLORINE MOLECULE Cl- This is what happens at thepositiveelectrode when chloride ions, Cl- are present in the electrolyte Here are negative chloride ions attracted towards the positive electrode.Opposite charges attract. As they get close, each Cl- ion loses an electron which goes onto the electrode. The ion becomes electrically neutral We have made chlorine the element. The neutral atoms join in pairs to make chlorine molecules, Cl2 which bubble off as a gas. 2Cl- - 2e  Cl2

  13. the ion GAINS electrons from the electrode Here are different ions that might be in a solution. You need to be able to balance the half equations. 2 2 click for solution click for solution H+ is present in all acids and hydrogen gas is evolved from the electrode. 2 2 4 2 2 2 In a special case, a positive copper electrode dissolves in a solution of copper sulphate. Electrons are lost by the copper metal. Half equations for reduction at the negative electrode. 3 2 Cu2+ + e  Cu Al3+ + e  Al 1 2 Ag+ + e  Ag Pb2+ + e  Pb In these cases, metallic elements would appear at the negative electrode. H+ + e  H2 Half equations for oxidation at the positive electrode. Cl-- e  Cl2 Br-- e  Br2 Here electrons are lost by the ion. When gaseous elements are produced, they bond together in pairs to make a molecule. The balancing needs to include this. 2 Cu - e  Cu2+ O2-- e  O2

  14. The end

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