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Which of these combinations will result in a reaction? Check all that apply. | $\square$ | $\mathrm{K}(\mathrm{s})+\mathrm{Au}^{3 +}(\mathrm{aq})$ | | --- | --- | | $\square$ | $\mathrm{K}^{+}(\mathrm{aq})+\mathrm{Au}(\mathrm{s})$ |
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Step 1:
: Identify the given half-reactions and their corresponding reduction potentials.

The first half-reaction is: \mathrm{K}(\mathrm{s}) \longrightarrow \mathrm{K}^{+}(\mathrm{aq}) + e^{-} \qquad E^{\circ} = - 2.93 \ \mathrm{V} The second half-reaction is: \mathrm{Au}^{3 +}(\mathrm{aq}) + 3 \ e^{-} \longrightarrow \mathrm{Au}(\mathrm{s}) \qquad E^{\circ} = 1.50 \ \mathrm{V}

Step 2:
: Multiply the number of electrons in each half-reaction by an integer so that the number of electrons is equal in both half-reactions.

In this case, we need to multiply the first half-reaction by 3 and the second half-reaction by 1. First half-reaction multiplied by 3: 3 \ \mathrm{K}(\mathrm{s}) \longrightarrow 3 \ \mathrm{K}^{+}(\mathrm{aq}) + 3 \ e^{-} \qquad E^{\circ} = - 2.93 \ \mathrm{V} Second half-reaction multiplied by 1: \mathrm{Au}^{3 +}(\mathrm{aq}) + 3 \ e^{-} \longrightarrow \mathrm{Au}(\mathrm{s}) \qquad E^{\circ} = 1.50 \ \mathrm{V}

Step 3:
: Add the two half-reactions together to get the overall balanced reaction.

Overall reaction: 3 \ \mathrm{K}(\mathrm{s}) + \mathrm{Au}^{3 +}(\mathrm{aq}) \longrightarrow 3 \ \mathrm{K}^{+}(\mathrm{aq}) + \mathrm{Au}(\mathrm{s})

Step 4:
: Determine if the reaction is spontaneous by calculating the standard cell potential, $E_{\text {cell}}^{\circ}$.

Since $E_{\text {cell}}^{\circ} > 0$, the reaction is spontaneous in the forward direction.
E_{\text {cell}}^{\circ} = E_{\text {cathode}}^{\circ} - E_{\text {anode}}^{\circ} = 1.50 \ \mathrm{V} - (- 2.93 \ \mathrm{V}) = 4.43 \ \mathrm{V}

Final Answer

The reaction between $\mathrm{K}(\mathrm{s})$ and $\mathrm{Au}^{3 +}(\mathrm{aq})$ will result in a reaction. The balanced reaction is: 3 \ \mathrm{K}(\mathrm{s}) + \mathrm{Au}^{3 +}(\mathrm{aq}) \longrightarrow 3 \ \mathrm{K}^{+}(\mathrm{aq}) + \mathrm{Au}(\mathrm{s})