Directions: Questions 1-3 are long free-response questions that require about 23 minutes each to answer and are worth 10 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. The unbalanced reaction between potassium permanganate and acidified iron (II) sulfate is a redox reaction that proceeds as follows: $$\mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-(a q)} \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$$ (a) Provide the equations for both half-reactions that occur below: (i) Oxidation half-reaction (ii) Reduction half-reaction (b) What is the balanced net ionic equation? A solution of 0.150 M potassium permanganate is placed in a buret before being titrated into a flask containing 50.00 mL of iron (II) sulfate solution of unknown concentration. The following data describes the colors of the various ions in solution: $$\begin{array}{|c|c|}\hline \text { Ion } & {\text { Color in solution }} \\\ \hline \mathrm{H^{+ }} & {\text { Colorless }} \\ \hline \mathrm{Fe}^{2+} & {\text { Pale Green }} \\ \hline \mathrm{MnO}_{4^{-}} & {\text {Dark Purple }} \\ \hline \mathrm{Mn}^{2+} & {\text { Colorless }} \\ \hline \mathrm{Fe}^{3+} & {\text { Yellow }} \\ \hline \mathrm{K}^{+} & {\text {Colorless }} \\ \hline \mathrm{SO}_{4}^{2-} & {\text { Colorless }} \\\ \hline\end{array}$$ (c) Describe the color of the solution in the flask at the following points: (i) Before titration begins (ii) During titration prior to the endpoint (iii) At the endpoint of the titration (d) (i) If 15.55 mL of permanganate are added to reach the endpoint, what is the initial concentration of the iron (II) sulfate? (ii) The actual concentration of the \(\mathrm{FeSO}_{4}\), is 0.250 \(M\) . Calculate the percent error. (e) Could the following errors have led to the experimental result deviating in the direction that it did? You must justify your answers quantitatively. (i) 55.0 \(\mathrm{mL}\) of \(\mathrm{FeSO}_{4}\) was added to the flask prior to titration instead of 50.0 mL . (ii) The concentration of the potassium permanganate was actually 0.160 \(M\) instead of 0.150 \(M\) .

Step by step solution

01

(a) Identifying Half-Reactions

The reactions that occur in a redox reaction are of two types: oxidation and reduction. (i) The oxidation half-reaction occurs where iron (II) is oxidized to iron (III): \(Fe^{2+} \rightarrow Fe^{3+} + e^-\). (ii) The reduction half-reaction occurs where permanganate ion is reduced to manganese (II) ion in acidic solution: \(MnO_{4}^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_{2}O\)

02

(b) Balanced Net Ionic Equation

The balanced net ionic equation is obtained by combining the half-reactions. We need to multiply the oxidation reaction by 5 and the reduction reaction by 1 to make sure the electrons lost will equal the electrons gained. The balanced redox equation then becomes: \(5Fe^{2+} + MnO_{4}^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_{2}O\)

03

(c) Colour Changes During Titration

(i) Before the titration begins, the solution in the flask contains Iron (II) which results in a pale green colour. (ii) During the titration process, the purple colour of the permanganate is reduced to a colourless solution. (iii) At the endpoint of the titration, All of Fe(II) has been oxidized to Fe(III), which is yellow, and so the solution turns yellow.

04

(d) (i) Determining the Initial Concentration of Iron (II) Sulfate

The moles of KMnO4 can be calculated using the formula moles = molarity x volume = 0.150 M x 15.55 ml/1000 = 0.0023325 mol. According to the balanced equation, 1 mol of permanganate ion reacts with 5 mol of Fe(II). Thus the moles of Fe(II) can be calculated as 5 times the moles of KMnO4 = 5 * 0.0023325 mol = 0.0116625 mol. Using the formula concentration = moles/volume, the original concentration of Fe(II) = 0.0116625 mol /50 ml = 0.23325 M

05

(d) (ii) Calculating the Percent Error

The percent error can be calculated using the formula: percent error = ( | experimental value - actual value | / actual value) x 100% = ( | 0.23325 - 0.250 | / 0.250) * 100% = 6.70%

06

(e) Evaluating Sources of Error

(i) If 55.0 mL of FeSO4 was added instead of 50.0 mL, it means more Fe(II) was used than planned. This would lead to a lower calculated concentration of Fe(II), as the concentration is inversely proportional to volume. (ii) If the concentration of MnO4- is 0.160M instead of 0.150, we would have needed less solution to reach the endpoint which would lead to a higher calculated concentration of Fe(II). With both errors, we see that they could have led to a calculated concentration that deviated from the actual value.

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