Question

Consider a voltaic cell in which the following reaction takes place. $$ 2 \mathrm{Fe}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O} $$ (a) Calculate \(E^{\circ}\). (b) Write the Nernst equation for the cell. (c) Calculate \(E\) at \(25^{\circ} \mathrm{C}\) under the following conditions: \(\left[\mathrm{Fe}^{2+}\right]=0.00813 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=0.914 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=0.199 \mathrm{M}, \mathrm{pH}=2.88\)

Short answer

Question: Calculate the cell potential under the given set of conditions for the following redox reaction: \( \mathrm{Fe}^{2+}(a q) + \mathrm{H}_{2} \mathrm{O}_{2}(a q) + 2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q) + 2 \mathrm{H}_{2} \mathrm{O} \) at \(25^{\circ} \mathrm{C}\), where the concentrations are: \([\mathrm{Fe}^{2+}]=0.00813 \mathrm{M}\), \([\mathrm{H}_{2}\mathrm{O}_{2}]=0.914 \mathrm{M}\), \([\mathrm{H}^{+}]=10^{-2.88}\mathrm{M}\), and \([\mathrm{Fe}^{3+}]=0.199 \mathrm{M}\). Answer: The cell potential under the given conditions is 1.65 V.

Step-by-step solution

Identify the half-reactions

To calculate the standard cell potential, it's essential to recognize and divide the given redox reaction into half-reactions. First, let's split the reaction into the oxidation and reduction half-reactions: Oxidation: \( \mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q) + \mathrm{e}^{-} \) Reduction: \( \mathrm{H}_{2} \mathrm{O}_{2}(a q) + 2 \mathrm{H}^{+}(a q) + 2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \)

Find the standard reduction potentials

In order to find the standard cell potential, we need to look up the standard reduction potentials for the half-reactions involved. Look up the following values in a standard reduction potential table: \(E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.77 \,\text{V}\) \(E^{\circ}_{H_2O_2/H_2O} = 1.77\,\text{V}\) Keep in mind these values are for reduction, so you may need to reverse the sign for oxidation.

Calculate the standard cell potential

We can calculate \(E^{\circ}\) using the standard reduction potentials: \(E^{\circ} = E^{\circ}_{\text{reduction}} - E^{\circ}_{\text{oxidation}}\) Since our oxidation half-reaction involves Fe²⁺, we'll need to reverse the sign of the standard potential for the Fe³⁺/Fe²⁺ couple: \(E^{\circ} = 1.77 - (-0.77) = 2.54\,\text{V}\)

Write the Nernst equation for the cell

The Nernst equation relates the cell potential to its standard cell potential and the concentrations of the species involved in the redox reaction: $$ E = E^{\circ} - \frac{0.0592}{n}\log_{10}{Q} $$ Where \(E^{\circ}\) is the standard cell potential, n is the number of electrons transferred in the redox reaction, and Q is the reaction quotient. In this case, n = 2 electrons and the Q can be expressed as: $$ Q = \frac{[\mathrm{Fe}^{3+}]^2[\mathrm{H}_{2}\mathrm{O}]^2}{[\mathrm{Fe}^{2+}]^2[\mathrm{H}_{2}\mathrm{O}_{2}][\mathrm{H}^{+}]^2} $$

Calculate the cell potential under given conditions

Use the concentrations and Nernst equation to calculate \(E\) at \(25^{\circ} \mathrm{C}\): $$ E = 2.54 - \frac{0.0592}{2}\log_{10}{\frac{(0.199)^2(1)^2}{(0.00813)^2(0.914)(10^{-2.88})^2}} $$ $$ E = 2.54 - (0.0296) \log_{10}{2283.37} $$ After solving for \(E\), we find: \(E = 1.65\,\text{V}\) So, the cell potential under the given conditions is 1.65 V.