Standard electrode potentials for two relevant half-reactions are: Cr2O7^2-(aq) + 14H^+(aq) + 6e^- 2Cr^3+(aq) + 7H2O(l) E^ = +1.33 V Fe^3+(aq) + e^- Fe^2+(aq) E^ = +0.77 V (a) Use these standard electrode potentials to show that the reaction between Cr2O7^2-(aq) and Fe^2+(aq) in…

A2 Chemistry 9701: Dichromate(VI)/Fe2+ Redox Feasibility and Titration Calculation

Standard electrode potentials for two relevant half-reactions are:

$\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightleftharpoons 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \qquad E^{\ominus} = +1.33\ \text{V}$ $\text{Fe}^{3+}(aq) + e^- \rightleftharpoons \text{Fe}^{2+}(aq) \qquad E^{\ominus} = +0.77\ \text{V}$

(a) Use these standard electrode potentials to show that the reaction between $\text{Cr}_2\text{O}_7^{2-}(aq)$ and $\text{Fe}^{2+}(aq)$ in acidic solution is thermodynamically feasible, and calculate the standard cell potential, $E^{\ominus}_{cell}$ , for this reaction. [2]

(b) Combine the two half-equations to write the overall ionic equation for this reaction. [2]

(c) A geologist analyses a $5.00\ \text{g}$ sample of crushed iron ore to determine its iron content. The sample is dissolved completely in excess dilute sulfuric acid and, by passing the resulting solution through a reducing column, all the iron present is converted to $\text{Fe}^{2+}(aq)$ ; this solution is then made up to exactly $250\ \text{cm}^3$ in a volumetric flask. A $25.0\ \text{cm}^3$ portion of this solution is titrated against $0.0150\ \text{mol dm}^{-3}$ standardised potassium dichromate(VI), using a few drops of sodium diphenylamine sulfonate as an indicator; $20.00\ \text{cm}^3$ is required to reach the end-point (a sharp colour change from green to violet). Calculate the number of moles of $\text{Cr}_2\text{O}_7^{2-}(aq)$ used, and hence the concentration, in $\text{mol dm}^{-3}$ , of $\text{Fe}^{2+}(aq)$ in the $25.0\ \text{cm}^3$ sample. [3]

(d) Hence calculate the percentage, by mass, of iron in the original ore sample. [molar mass of $\text{Fe}=55.8\ \text{g mol}^{-1}$ ] [2]

Part (a): Feasibility and the standard cell potential

The half-equation with the more positive $E^{\ominus}$ has the greater tendency to proceed in the direction of reduction (as written), and so takes electrons from the other half-cell. Since $E^{\ominus}(\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+})=+1.33\ \text{V} > E^{\ominus}(\text{Fe}^{3+}/\text{Fe}^{2+})=+0.77\ \text{V},$ $\text{Cr}_2\text{O}_7^{2-}(aq)$ is reduced to $\text{Cr}^{3+}(aq)$ , while $\text{Fe}^{2+}(aq)$ is oxidised to $\text{Fe}^{3+}(aq)$ .

$E^{\ominus}_{cell}=E^{\ominus}_{reduction}-E^{\ominus}_{oxidation}=1.33-0.77=+0.56\ \text{V}$

Since $E^{\ominus}_{cell}$ is positive, the reaction between $\text{Cr}_2\text{O}_7^{2-}(aq)$ and $\text{Fe}^{2+}(aq)$ in acidic solution is thermodynamically feasible.

Part (b): Combining the half-equations

The reduction half-equation transfers 6 electrons; the oxidation half-equation transfers only 1. Multiplying the iron half-equation by 6 so the electrons cancel: $\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l)$ $6\text{Fe}^{2+}(aq) \rightarrow 6\text{Fe}^{3+}(aq) + 6e^-$

Adding these and cancelling the $6e^-$ on each side gives the overall ionic equation: $\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6\text{Fe}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) + 6\text{Fe}^{3+}(aq)$

(Charge check: left-hand side $-2+14+12=+24$ ; right-hand side $+6+0+18=+24$ . Balanced.)

Part (c): Moles of $\text{Cr}_2\text{O}_7^{2-}$ and concentration of $\text{Fe}^{2+}$

Moles of $\text{Cr}_2\text{O}_7^{2-}(aq)$ used: $n(\text{Cr}_2\text{O}_7^{2-}) = 0.0150\ \text{mol dm}^{-3} \times \frac{20.00}{1000}\ \text{dm}^3 = 3.00\times10^{-4}\ \text{mol}$

From the equation in (b), the mole ratio $\text{Cr}_2\text{O}_7^{2-}:\text{Fe}^{2+}$ is $1:6$ , so: $n(\text{Fe}^{2+}) = 6 \times 3.00\times10^{-4} = 1.80\times10^{-3}\ \text{mol}$

This is the amount of $\text{Fe}^{2+}$ present in the $25.0\ \text{cm}^3$ sample, so: $[\text{Fe}^{2+}] = \frac{1.80\times10^{-3}\ \text{mol}}{0.0250\ \text{dm}^3} = 0.0720\ \text{mol dm}^{-3}$

Part (d): Percentage of iron in the ore

The $25.0\ \text{cm}^3$ sample was pipetted from the full $250\ \text{cm}^3$ volumetric flask (a factor of $10$ scale-up). Moles of $\text{Fe}^{2+}$ in the whole flask: $n(\text{Fe}^{2+})_{total} = 0.0720\ \text{mol dm}^{-3} \times 0.250\ \text{dm}^3 = 1.80\times10^{-2}\ \text{mol}$

Since all the iron in the ore was converted to $\text{Fe}^{2+}(aq)$ , this is also the total moles of iron in the sample. Mass of iron: $m(\text{Fe}) = 1.80\times10^{-2}\ \text{mol} \times 55.8\ \text{g mol}^{-1} = 1.0044\ \text{g}$

(Check: $0.0180\times55.8=0.0180\times50+0.0180\times5.8=0.900+0.1044=1.0044\ \text{g}$ . Consistent.)

Percentage of iron, by mass, in the $5.00\ \text{g}$ ore sample: $\%\ \text{Fe} = \frac{1.0044}{5.00}\times100 = 20.088\% \approx 20.1\%\ (3\ \text{s.f.})$

Final answers

(a) $E^{\ominus}_{cell}=+0.56\ \text{V}$ . Positive, so the reaction is feasible.
(b) $\text{Cr}_2\text{O}_7^{2-}(aq)+14\text{H}^+(aq)+6\text{Fe}^{2+}(aq)\rightarrow2\text{Cr}^{3+}(aq)+7\text{H}_2\text{O}(l)+6\text{Fe}^{3+}(aq)$
(c) $n(\text{Cr}_2\text{O}_7^{2-})=3.00\times10^{-4}\ \text{mol}$ ; $[\text{Fe}^{2+}]=0.0720\ \text{mol dm}^{-3}$
(d) Percentage of iron in the ore $\approx20.1\%$ (3 s.f.)