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Chapter 18: Problem 119

Hydrogen peroxide can function either as an oxidizing agent or as a reducing agent. At standard conditions, is \(\mathrm{H}_{2} \mathrm{O}_{2}\) a better oxidizing agent or reducing agent? Explain.

Short Answer

Expert verified
At standard conditions, hydrogen peroxide (H2O2) is a better oxidizing agent than a reducing agent. This is because its standard reduction potential (Eº) for the reduction half-reaction (+1.77 V) is greater than that of the oxidation half-reaction (+0.68 V), indicating a higher tendency to undergo reduction and thus act as an oxidizing agent.

Step by step solution

01

Write the half-reactions

To analyze H2O2's behavior as an oxidizing or reducing agent, we first need to write the redox half-reactions for its oxidation and reduction. Oxidation half-reaction (H2O2 acts as a reducing agent): \[ \mathrm{H}_{2}\mathrm{O}_{2} \to \mathrm{O}_{2} + 2\mathrm{H}^{+} + 2\mathrm{e}^{-} \] Reduction half-reaction (H2O2 acts as an oxidizing agent): \[ \mathrm{H}_{2}\mathrm{O}_{2} + 2\mathrm{H}^{+} + 2\mathrm{e}^{-} \to 2\mathrm{H}_{2}\mathrm{O} \]
02

Find the standard reduction potentials (Eº) for the half-reactions

Next, we need to find the standard reduction potentials (Eº) for these half-reactions. Eº values can be found in standard reduction potential tables. For the oxidation half-reaction (H2O2 acts as a reducing agent): Eº = +0.68 V For the reduction half-reaction (H2O2 acts as an oxidizing agent): Eº = +1.77 V
03

Compare the Eº values to determine if H2O2 is a better oxidizing or reducing agent

Now, we can compare the Eº values for both half-reactions to identify if H2O2 is a better oxidizing or reducing agent. As a general rule, the more positive the Eº value, the greater the tendency for the substance to undergo reduction (and thus act as an oxidizing agent). Since +1.77 V (reduction half-reaction) is greater than +0.68 V (oxidation half-reaction), H2O2 has a higher tendency to undergo reduction. With the given information, we can now conclude that at standard conditions, hydrogen peroxide (H2O2) is a better oxidizing agent than a reducing agent.

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Most popular questions from this chapter

Three electrochemical cells were connected in series so that the same quantity of electrical current passes through all three cells. In the first cell, 1.15 g chromium metal was deposited from a chromium (III) nitrate solution. In the second cell, 3.15 \(\mathrm{g}\) osmium was deposited from a solution made of \(\mathrm{Os}^{n+}\) and nitrate ions. What is the name of the salt? In the third cell, the electrical charge passed through a solution containing \(\mathrm{X}^{2+}\) ions caused deposition of 2.11 \(\mathrm{g}\) metallic \(\mathrm{X}\) . What is the electron configuration of \(\mathrm{X} ?\)

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\\ \hline{M^{4+}+4 e^{-} \longrightarrow M} & {0.66} \\ {N^{3+}+3 e^{-} \longrightarrow N} & {0.39}\end{array}$$ What is the value of \(\Delta G^{\circ}\) and \(K\) for this cell?

An electrochemical cell is set up using the following unbalanced reaction: $$\mathrm{M}^{a+}(a q)+\mathrm{N}(s) \longrightarrow \mathrm{N}^{2+}(a q)+\mathrm{M}(s)$$ The standard reduction potentials are: $$\mathrm{M}^{a+}+a \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=0.400 \mathrm{V}$$ $$\mathrm{N}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{N} \quad \mathscr{E}^{\circ}=0.240 \mathrm{V}$$ The cell contains 0.10\(M \mathrm{N}^{2+}\) and produces a voltage of 0.180 \(\mathrm{V}\) . If the concentration of \(\mathrm{M}^{a+}\) is such that the value of the reaction quotient \(Q\) is \(9.32 \times 10^{-3},\) calculate \(\left[\mathrm{M}^{a+}\right] .\) Calculate \(w_{\text { max }}\) for this electrochemical cell.

Consider the following half-reactions: $$\begin{array}{ll}{\mathrm{IrCl}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.77 \mathrm{V}} \\\ {\mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.73 \mathrm{V}} \\\ {\mathrm{PdCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-}} & {\mathscr{E}^{\circ}=0.62 \mathrm{V}}\end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0\(M\) in chloride ion and 0.020\(M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that 99\(\%\) of a metal must be plated out before another metal begins to plate out.)

Consider the cell described below: $$\text { Al }\left|\mathrm{Al}^{3+}(1.00 M)\right|\left|\mathrm{Pb}^{2+}(1.00 M)\right| \mathrm{Pb}$$ Calculate the cell potential after the reaction has operated long enough for the [Al \(^{3+} ]\) to have changed by 0.60 \(\mathrm{mol} / \mathrm{L}\) . (Assume \(T=25^{\circ} \mathrm{C} .\)
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