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Chapter 17: Problem 59

Calculate the voltage of a cell that has the following electrode reactions: \begin{equation} \begin{array}{c}{2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2}(g)} \\ {\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)+e^{-}}\end{array} \end{equation}

Short Answer

Expert verified
The voltage of the cell is 0.77 V

Step by step solution

01

Identify the Reduction and Oxidation Reactions

From the provided reactions:\n - The first one: \(2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+2 e^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2}(g)\) is the reduction reaction as it involves the gain of electrons (e-).\n - The second one: \(\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{Fe}^{3+}(a q)+e^{-}\) is the oxidation reaction as it involves the loss of an electron.
02

Determine the Individual Electrode Potentials

You will need an Electrochemical Series chart to find the individual electrode potentials. The reduction potential for the hydrogen ion (\(H_3O^+\)) to \(H_2(g)\) is 0 V (because the \(H^+/H_2\) half-cell is the standard for potential difference) and the standard oxidation potential for \(Fe^{2+}\) to \(Fe^{3+}\) is 0.77 V.
03

Calculate the Cell potential

Sum the standard reduction potential and the standard oxidation potentials to get the cell potential (Ecell) - which is \(E_{cell} = E_{oxidation} + E_{reduction}\). Substitute the known values: \(E_{cell} = 0.77 V + 0 V = 0.77 V\)

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