Given the standard electrode potentials, K⁺/K = -2.93V, Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Mg²⁺/Mg = -2.37V, Cr³⁺/Cr = -0.74V. Arrange these metals in their increasing order of reducing power.
The increasing order of reducing power is: Ag < Hg < Cr < Mg < K.
According to the standard electrode potentials provided, metals with lower (more negative) electrode potentials have higher reducing power. The electrode potentials are: Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Cr³⁺/Cr = -0.74V, Mg²⁺/Mg = -2.37V, and K⁺/K = -2.93V.
As stated in the text, as the standard electrode potential decreases, the reducing power of the species on the right side of the reaction increases. Therefore, K with the lowest electrode potential (-2.93V) has the highest reducing power, while Ag with the highest electrode potential (0.80V) has the lowest reducing power.
Explanation
The textbook clearly establishes the relationship between standard electrode potential and reducing power: as electrode potential decreases (becomes more negative), the reducing power increases. This is because a lower reduction potential means the metal has a greater tendency to lose electrons and get oxidized, making it a stronger reducing agent. The question requires arranging metals from weakest to strongest reducing agent, which corresponds to arranging them from highest to lowest electrode potential.
Solution Steps
Step 1: List the given electrode potentials: Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Cr³⁺/Cr = -0.74V, Mg²⁺/Mg = -2.37V, K⁺/K = -2.93V
Step 2: Apply the principle that lower electrode potential means higher reducing power
Step 3: Arrange metals from highest to lowest electrode potential to get increasing order of reducing power: Ag (0.80V) < Hg (0.79V) < Cr (-0.74V) < Mg (-2.37V) < K (-2.93V)