You Consider the following reaction at \( 298 \mathrm{~K} \). \[ \begin{array}{l} \mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow \mathrm{H}_{2}(g)+\mathrm{Fe}^{3+}(a q) \\ \begin{array}{|l|l|} \hline \mathrm{H}^{+}(a q)+2 e^{-} \rightarrow \mathrm{H}_{2}(g) & E_{\text {red }}^{o}=0.000 \mathrm{~V} \\ \hline \mathrm{Fe}^{3+}(a q)+e^{-} \rightarrow \mathrm{Fe}^{2+}(a q) & E_{\text {red }}^{o}=0.771 \mathrm{~V} \\ \hline \end{array} \end{array} \] Which of the following statements are correct? (Select all that apply.) \( \Delta G^{0}>0 \) \( n=2 \) mol electrons The reaction is product-favored. \( K<1 \) \( E_{\text {cell }}^{\circ}>0 \)
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The given reactions are: H+(aq) + 2e- → H2(g) E°red = 0.000 V Fe3+(aq) + e- → Fe2+(aq) E°red = 0.771 V We need to reverse the second reaction to match the overall reaction: Fe2+(aq) → Fe3+(aq) + e- E°red = -0.771 V Then, we add the two reactions Show more…
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Determine $E_{\text {cell }}^{\circ}$ and $\Delta G^{\circ}(298 \mathrm{K})$ for the following reactions: (a) $\mathrm{Co}^{2+}(\mathrm{aq})+2 \mathrm{Cr}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Co}(\mathrm{s})+2 \mathrm{Cr}^{3+}(\mathrm{aq})$ (b) $\frac{1}{2} \mathrm{Co}^{2+}(\mathrm{aq})+\mathrm{Cr}^{2+}(\mathrm{aq}) \rightleftharpoons \frac{1}{2} \mathrm{Co}(\mathrm{s})+\mathrm{Cr}^{3+}(\mathrm{aq})$ (c) $\left[\mathrm{Cr}_{2} \mathrm{O}_{7}\right]^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}(\mathrm{aq})+6 \mathrm{Fe}^{2+}(\mathrm{aq}) \rightleftharpoons$ $2 \mathrm{Cr}^{3+}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+6 \mathrm{Fe}^{3+}(\mathrm{aq})$
Given the following half-reactions, combine the two that give the cell reaction with the most positive $E^{\circ} .$ Write a balanced equation for the cell reaction, and calculate $E^{\circ}$ and $\Delta G^{\circ} .$ $$\begin{array}{ll}{\mathrm{Co}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}(s)} & {E^{\circ}=-0.28 \mathrm{V}} \\ {\mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}(a q)} & {E^{\circ}=0.54 \mathrm{V}} \\ {\mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)} & {E^{\circ}=0.34 \mathrm{V}}\end{array}$$
Use Table 17.1 to answer the following questions. Use LT (for is less than), GT (for is greater than), EQ (for is equal to), or MI (for more information required). (a) For the half reaction $$ \begin{aligned} & \frac{1}{2} \mathrm{Br}_{2}(l)+e^{-} \longrightarrow \mathrm{Br}^{-}(a q) \\ E_{\mathrm{red}}^{\circ} & 1.077 \mathrm{~V} \end{aligned} $$ (b) For the reaction $$ 2 \mathrm{Br}^{-}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow \mathrm{Br}_{2}(l)+\mathrm{Co}(s) $$ $E^{\circ}$ $0 .$ (c) If the half reaction $$ \mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s) $$ is designated as the new standard where $E_{\text {red }}^{\circ}$ is $0.00,$ then $E_{\text {red }}^{\circ}$ for $$ 2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g) $$ is $\quad 0 .$ (d) For the reaction $$ 2 \mathrm{Cr}^{3+}(a q)+3 \mathrm{Co}(s) \longrightarrow 2 \mathrm{Cr}(s)+3 \mathrm{Co}^{2+}(a q) $$ the number of electrons exchanged is (e) For the reaction described in (d), the number of coulombs that passes through the cell is $9.648 \times 10^{4}$
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