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Working Out Formulae & Balancing Equations (including electrolysis) Valency Using Valency Writing Equations Balancing Equations END ALWAYS BRINGS YOU.

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Presentation on theme: "Working Out Formulae & Balancing Equations (including electrolysis) Valency Using Valency Writing Equations Balancing Equations END ALWAYS BRINGS YOU."— Presentation transcript:

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

3 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 ElementGroupOuter ElectronsValency Sodium111 Chlorine778 – 7 = 1 Oxygen668 – 6 = 2 Magnesium222 Nitrogen558 – 5 = 3 Carbon444

4 Here’s another method that works for ions… For IONIC BONDS, the valency is just the size of the charge on the ion Sodium forms Na + ionIt has valency 1 Calcium forms Ca 2+ ionIt has valency 2 Oxygen forms O 2  ionIt has valency 2 Chlorine forms Cl  ionIt has valency 1 Some transition metals can form more than one ion, so they have more than one valency. Iron forms Fe 2+ and Fe 3+ So it can have valency 2 or 3 You can use the same idea for ions made of more than one atom Ammonium ion NH 4 + has valency 1 Nitrate ion NO 3  has valency 1 Sulphate ion SO 4 2  has valency 2

5 USING VALENCIES TO WORK OUT FORMULAE 1.Write down the two elements (or ions) in the compound 2.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. 3.Cancel if necessary, and remove any 1s Sodium sulphate Na SO 4 Valency of sodium: 1 Valency of sulphate: 2 So Na 2 (SO 4 ) 1 Remove 1s: Na 2 SO 4 Calcium oxide: Ca O Valency of calcium: 2 Valency of oxygen: 2 So Ca 2 O 2 Cancel: CaO Iron(II) chloride Fe Cl Valency of iron : 2 Valency of chloride : 1 So Fe 1 Cl 2 Remove 1s: Fe Cl 2 DO NOT “MULTIPLY OUT THE BRACKETS” ! Write (NO 3 ) 2 not N 2 O 6 Magnesium nitrate Mg NO 3 Valency of magnesium : 2 Valency of nitrate : 1 So Mg 1 (NO 3 ) 2 Remove 1s: Mg (NO 3 ) 2

6 WRITING EQUATIONS Step 2: Write down the FORMULA of each of the chemicals in your word equation The chemicals you start with What you make in the reaction Reactants Products Step 1: Write a 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 REACTANTS calcium carbonate hydrochloric acid + calcium chloride + water + carbon dioxide The PRODUCTS CaCO 3 + HCl CaCl 2 + H 2 O + CO 2 Make sure you get the formulae right! No marks for balancing if the formulae are wrong! (s) (aq) (l) (g) Calcium carbonate is an insoluble solid Hydrochloric acid, like other acids, is always a solution 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 Now on to the harder bit – balancing the equation!

7 They all match. So it’s balanced! Balancing Equations 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 + 2H 2 O  2NaOH + H 2 AtomNumber on LeftNumber on Right NaHave 2Na, so 2 H 2 in each water 2 waters 2  2 = 4 2 from 2NaOH + 2 from H 2 makes 4 O 1 in each water 2 waters. So 2 1 in each NaOH 2NaOH. So 2 1 in each NaOH 2NaOH. So 2

8 NaOH + H 2 SO 4  Na 2 SO 4 + H 2 O How to balance an equation 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! There are different numbers of Na and H Na is only in one chemical each side We can balance them by putting 2 in front of NaOH 2 Now recalculate the numbers of atoms… LeftRight Na12 O55 H32 S There are different numbers of O and H H is in fewer different formulae. We can balance H by putting 2 in front of H 2 O 2 Now recalculate the numbers of atoms… 4 6 THEY ALL MATCH! IT’S BALANCED!

9 Other examples… LeftRight Ca11 C 1 1 O33 H12 Cl12 CaCO 3 + HCl  CaCl 2 + H 2 O + CO Let’s balance the hydrogens: We can do this by putting 2 in front of HCl 2 Eureka ! That has automatically balanced the chlorines too. balanced LeftRight Cl21 K 1 1 I12 Cl 2 + KI  KCl + I 2 2 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 2 2 balanced

10 Another example… Al 2 O 3 + HCl  AlCl 3 + H 2 O None of the atoms are balanced! They all occur in just two chemicals Choose one to balance… 23 We can balance Al by putting 2 in front of AlCl 3 Recalculate… Now let’s balance the oxygens: We can do this by putting 3 in front of H 2 O Recalculate… 6 Now let’s balance the hydrogens: We can do this by putting 6 in front of HCl Recalculate… THEY ALL MATCH! IT’S BALANCED! LeftRight Al21 O31 H12 Cl

11 An awkward one! Al + Cl 2  AlCl 3 Chlorines aren’t balanced. 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 To do that, we put 3 in front of Cl 2 and 2 in front of AlCl But how can we do the balancing? We haven’t got “nice” numbers! Recalculate… Now we must balance the aluminiums We can do this by putting 2 in front of Al LeftRight Al11 Cl Recalculate… 2 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

12 Cu 2+ As they get close, the ions gain electrons from the electrode and the Cu 2+ is neutralised. Here are Cu 2+ ions moving to the negative electrode. Positive ions in the solution are attracted to negative electrode – opposite charges attract. The electrode is negative because it has too many electrons e - go to ion. 2e - + Cu 2+  Cu 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 This makes copper the element, which covers the electrode. e - go to ion TWO ELECTRONS FROM THE CATHODE ARE ADDED TO THE COPPER ION A NEUTRAL ATOM OF THE ELEMENT COPPER. ELECTROLYSIS

13 This is what happens at the positive electrode when chloride ions, Cl - are present in the electrolyte This electrode is positive because some electrons have been removed by the cell. 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 the ion loses an e - the ion loses an e - Cl - Cl 2 e - go to cell We have made chlorine the element. The neutral atoms join in pairs to make chlorine molecules, Cl 2 which bubble off as a gas. 2Cl - - 2e  Cl 2 TWO CHLORIDE IONS, EACH WITH AN EXTRA ELECTRON THE TWO ELECTRONS LOST BY THE IONS GO TO THE ELECTRODE A NEUTRAL CHLORINE MOLECULE 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

14 Half equations for reduction at the negative electrode. Cu 2+ + e  Cu click for solution 2 Al 3+ + e  Al 3 Ag + + e  Ag 1 Pb 2+ + e  Pb 2 H + + e  H 2 click for solution H + is present in all acids and hydrogen gas is evolved from the electrode. 2 2 In these cases, metallic elements would appear at the negative electrode. Half equations for oxidation at the positive electrode. 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. Cl - - e  Cl 2 22 Br - - e  Br 2 22 O e  O 2 24 Cu - e  Cu 2+ In a special case, a positive copper electrode dissolves in a solution of copper sulphate. Electrons are lost by the copper metal. 2 Here are different ions that might be in a solution. You need to be able to balance the half equations. the ion GAINS electrons from the electrode

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