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Hello.
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In this question, we're given six different equations and were asked to balance them using two different methods.
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One is the oxidation number method and the other is the half reaction method.
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First, let's start with the oxidation number method.
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Let's look at equation 8.
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We are given f .e plus 2 plus mn 04 minus, giving fe plus 3 plus mn plus 2.
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And this takes place in acidic conditions.
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Now let's first assign the oxidation numbers.
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So we have plus 2.
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Mn 04 minus the mn has a plus 7.
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Fe plus 3 and plus 2.
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Now let's see the change in oxidation number.
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So fe plus 2 to fe plus 3 there's a change in oxidation number involving one electron.
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And an mn plus 7 to mn plus 2, there's a change in oxidation for 5.
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Electrons.
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Now, since we need to balance these electrons, we'll be multiplying this by 5.
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Hence, the fe plus 2 and fe plus 3 will obtain the coefficient of 5.
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So now let's write the equation with balanced electrons.
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So we have 5 fe plus 2 plus mno4 minus giving 5 fe plus 3 plus mn plus 2.
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Now we need to balance the oxygens.
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Since on the left side, we have have 4 oxygens, we need to balance it on the right side by adding 4 h2o...