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The balanced chemical equation for silver reacting with sulfuric acid is: $$2\mathrm{Ag(s)} + 2\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \rightarrow \mathrm{Ag}_{2}\mathrm{SO}_{4}(s) + 2\mathrm{H}_{2}(g)$$

For comparison purposes, we can also write down the balanced chemical equation for the reaction of gold with sulfuric acid, even though it doesn't actually occur: $$2\mathrm{Au(s)} + 3\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \rightarrow \mathrm{Au}_{2}(\mathrm{SO}_{4})_{3}(s) + \mathrm{3H}_{2}(g)$$

A reducing agent is a substance that donates (loses) electrons to another substance in a redox reaction. Let's examine the redox behavior of silver and gold in their respective reactions by identifying the oxidation states of each element before and after the reaction: For silver: - In elemental form, silver has an oxidation state of 0. - In silver sulfate (Ag2SO4), silver has an oxidation state of +1. For gold: - In elemental form, gold has an oxidation state of 0. - In gold sulfate (Au2(SO4)3), gold would have an oxidation state of +3 (if the reaction happened).

Since a reducing agent loses electrons in the redox reaction, we can look at how the oxidation state of each metal changes to assess their reducing ability: - Silver goes from an oxidation state of 0 to +1, meaning it loses one electron per atom, reacting with sulfuric acid. - Gold would go from an oxidation state of 0 to +3 if it were to react with sulfuric acid, meaning it would lose three electrons per atom. Since gold doesn't dissolve in sulfuric acid, this indicates that gold is not willing to lose three electrons in the reaction. Therefore, silver, which dissolves in sulfuric acid and forms silver sulfate, is the better reducing agent because it's more willing to lose electrons and reduce other substances.