2. Given the following reaction between iron(II) with permanganate ion in an aci

Question

2. Given the following reaction between iron(II) with permanganate ion in an acidic solution. Which atoms experience a change in oxidation number? Identify which compound is reduced (oxidising agent) and which compound is oxidized. (reducing agent) 3. Calculate the grams of iron ion present in 12.5 mL of a solution that required 2.20 mL of 0.02115 M KMnO4 in a titration going to a fait pink endpoint. 4. In acid-base titration an indicator is added. Why is an indicator not used in the redox titrations from this experiment?

Explanation / Answer

2)

5Fe+2 + MnO4- + 8H- ----> 5Fe+3 +Mn-2 + 4H2O

Atoms changing oxidation number:

Fe+2 to Fe+3

Mn+7 to Mn-2

H-1 to H+1

Mn(7+) is reduced to Mn2+

Fe+2 is being oxidized
the oxidizing agent is MnO4-1
The reducing agent is Fe2+

3) Grams of Iron in of solution required

V1 = 1.25 ml

V2 = 2.20 ml

M2 = 0.02115 KMn4O

From the previous exercise:

5 mol of Fe+2 react with 1 mol of MnO4-

Therefore

Moles of MnO4- = M2*V2

moles = M2*V2 = 2.25 ml * 0.02115 = 0.0475 mmol of MnO4-

We need 5 moles of for every 1 mol of MnO4- therefore, 5x(0.0475) = 0.2375 moles of Fe+2

Since MW of Fe = 55 g/gmol then

mol of Fe+2 = 55g/gmol * 0.2375 mol = 13.06 grams of Iron present in those1.25 ml

Mass of Iron = 13.06

3) In acid-base there is a change in pH and the indicator will change of color becaus bases/acids rarely change their color solution. In this case, the Mixture of solution will change by itself Mn4O- will change to pink (Mn-)

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