Balancing Redox Equations

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Take permanganate ions in acid, and add iron two plus ions. Each permanganate ion reacts with a fixed number of iron ions. One? Two? No. Exactly five. But why five? Interesting, right? By the end of this video, you will get that five yourself. Balancing redox equations comes up often in JEE, so let's begin.

0:30The ion–electron method (acidic medium)

So, how do we balance a redox equation? We use the ion electron method, also called the half-reaction method. Step one: split the reaction into two halves, one for the oxidation, the loss of electrons, and one for the reduction, the gain of electrons. Okay? Step two: in each half, balance all the atoms other than oxygen and hydrogen. Those two we keep for later. Right? Now, suppose one side has fewer oxygen atoms than the other. Remember, the reaction happens in water. So what can we add to fix the oxygen? Water, yes, exactly! Add H two O to the side that is short of oxygen. The small two below is a subscript: two hydrogens in each water. But now the hydrogen is off. So, in acidic medium, balance it with H plus ions; the plus sign up top is the charge. Clear? Now the atoms are balanced, but the charges on the two sides still differ. Electrons are negative. So which side should they go on? The side with more positive charge, correct! Add just enough electrons there to make the charges equal. On screen, e with a small minus sign up top is one electron. So in an oxidation half, the electrons appear on the right; in a reduction half, on the left. Got it? Last step. Electrons cannot be left over: whatever one half gives, the other must take. So multiply the halves until the electrons are equal, then add, and the electrons cancel. With me?

2:02Basic medium: one extra step

Now, what if the reaction is in basic medium? The method is the same. First balance everything exactly as in acid, with H plus ions. Okay? But a basic solution is full of O H minus, the hydroxide ion, so H plus cannot stay in the final answer. So, count the H plus ions, and add that many O H minus ions to both sides. Right? Now, a common slip. Some students add O H minus only on the side that has the H plus. What goes wrong then? Yes, the equation is no longer balanced! Add the same thing to both sides, and it stays balanced. Then each H plus and O H minus pair joins to make one water; the arrow means gives. If water appears on both sides, cancel the extra. Clear?

2:54The oxidation-number method

There is a second method too, the oxidation number method. First, find which elements change their oxidation number, and by how much. One goes up, another comes down. Why must the total increase equal the total decrease? Because the electrons lost must equal the electrons gained, exactly! So choose the numbers in front to make the total increase equal to the total decrease. Got it? Then balance the oxygen and the hydrogen at the end. We will use this method as a quick check in example one.

3:27Example 1: MnO4- with Fe2+ in acid, why five Fe2+

Now, example one, our puzzle from the start. Balance M n O four minus plus F e two plus, giving M n two plus and F e three plus, in acid. M n O four minus is the permanganate ion. Pause the video and try it first. Done? Okay, let's check. Reduction half first: permanganate goes to M n two plus. Manganese is already balanced. But there are four oxygens on the left, and none on the right. So how many water molecules do we add on the right? Four, correct! Four waters bring eight hydrogens, so add eight H plus on the left. Now the charge: minus one plus eight is plus seven on the left, and plus two on the right. So add five electrons on the left, and both sides become plus two. Okay? Now the oxidation half. F e two plus gives F e three plus, and one electron. But wait. This half gives one electron, and the manganese half takes five. So what do we do? Multiply the iron half by five, yes! The cross sign means times. Now five electrons are given and five are taken. And look, the trap: add the halves without multiplying, and electrons are left over. That is never right. With me? Now add the two halves, and the five electrons cancel. M n O four minus, plus five F e two plus, plus eight H plus, gives M n two plus, plus five F e three plus, plus four waters. One permanganate for exactly five iron ions. Puzzle solved! Now a quick check with the oxidation number method. Manganese goes from plus seven to plus two: it falls by five. Each iron goes from plus two to plus three: it rises by one. So five irons for one manganese. Same answer! Right? And the charge: on the left, minus one plus ten plus eight gives plus seventeen. On the right, plus two plus fifteen, also plus seventeen. Perfect!

5:37Example 2: MnO4- with I- in basic medium (MnO2, IO3-)

Now a harder one, example two, and this time in basic medium. Balance M n O four minus plus I minus, giving M n O two and I O three minus. I minus is the iodide ion, and I O three minus is the iodate ion. Pause and try it. Hint: balance it as in acid first. Done? Okay, let's check. Permanganate goes to M n O two: four oxygens on the left, two on the right, so two waters on the right. That gives four hydrogens, so four H plus on the left. Charge: plus three on the left, zero on the right, so three electrons on the left. Clear? Oxidation half: iodide goes to iodate. Iodate has three oxygens, so three waters on the left. Six hydrogens now, so six H plus on the right. Charge: minus one on the left, plus five on the right, so six electrons on the right. Okay? So the reduction half takes three electrons, and the oxidation half gives six. How do we make them equal? Multiply the reduction half by two, exactly! Now six electrons are taken and six are given. Got it? Now add. Eight H plus on the left and six on the right, so two stay on the left. Four waters on the right and three on the left, so one stays on the right. Two M n O four minus, plus I minus, plus two H plus, gives two M n O two, plus I O three minus, plus one water. Atoms balanced, charge balanced. So we are done, right? No, no, careful! This is a common JEE trap. The question says basic medium, and we still have two H plus. So add two O H minus to both sides. On the left, two H plus and two O H minus become two waters. With me? Now two waters on the left and one on the right, so cancel one from each side. Two M n O four minus, plus I minus, plus one water, gives two M n O two, plus I O three minus, plus two O H minus. Charge check: minus three on each side. That's it!

7:57Quick revision: halves, electrons, OH- in basic medium

So, quick revision. Split the reaction into two halves. Balance oxygen with water, hydrogen with H plus, and the charge with electrons. Multiply so that the electrons cancel, then add. In basic medium, add O H minus to both sides, then cancel water. And check: the total rise in oxidation number equals the total fall. Now go and try the practice problems on this page. All the best!