Question

Suppose \(E_{\text {red }}^{\circ}\) for \(\mathrm{Ag}^{+} \longrightarrow\) Ag were set equal to zero instead of that of \(\mathrm{H}^{+} \longrightarrow \mathrm{H}_{2} .\) What would be (a) \(E_{\text {red }}^{\circ}\) for \(\mathrm{H}^{+} \longrightarrow \mathrm{H}_{2}\) ? (b) \(E_{\mathrm{ox}}^{\circ}\) for \(\mathrm{Ca} \longrightarrow \mathrm{Ca}^{2+} ?\) (c) \(E^{\circ}\) for the cell in \(33(\mathrm{c})\) ? Compare your answer with that obtained in \(33(\mathrm{c})\).

Short answer

Answer: The new standard reduction potential for H⁺ → H₂ is -0.80 V, the new standard oxidation potential for Ca → Ca²⁺ is 3.67 V, and the standard cell potential for the new cell remains the same at 0.80 V.

Step-by-step solution

(a) Standard reduction potential for H⁺ → H₂

Since the standard reduction potential for Ag⁺ → Ag is set to zero, we first need to find the new value E° for H⁺ → H₂. Given the original standard reduction potential for Ag⁺ → Ag (E°(Ag⁺/Ag) = 0.80 V) and H⁺ → H₂ (E°(H⁺/H₂) = 0 V), the difference in their E° values is 0.80 V (E°(Ag⁺/Ag) - E°(H⁺/H₂) = 0.80). Now, if we set E°(Ag⁺/Ag) = 0 V, we'll have E°(H⁺/H₂) = -0.80 V. Hence, the new standard reduction potential for H⁺ → H₂ will be -0.80 V.

(b) Standard oxidation potential for Ca → Ca²⁺

Given the original standard reduction potential for Ca²⁺ → Ca, E°(Ca²⁺/Ca) = -2.87 V. We can find the standard oxidation potential (E°(Ca/Ca²⁺)) by changing the sign of the reduction potential: E°(Ca/Ca²⁺) = -E°(Ca²⁺/Ca) = 2.87 V. The difference between the new standard reduction potentials of H⁺ → H₂ and Ag⁺ → Ag is 0.80 V (as calculated in (a)). Thus, we need to adjust the standard oxidation potential of Ca → Ca²⁺ accordingly. New E°(Ca/Ca²⁺) = E°(Ca/Ca²⁺) + difference = 2.87 V + 0.80 V = 3.67 V.

(c) Standard cell potential for the new cell

For this part, let's consider the cell reaction from exercise 33(c): H₂(g) + 2Ag⁺ → 2H⁺ + 2Ag(s) The original half-cell reactions and their standard reduction potentials are: Ag⁺ + e⁻ → Ag (E°(Ag⁺/Ag) = 0.80 V) 2H⁺ + 2e⁻ → H₂ (E°(H⁺/H₂) = 0 V) While the new values are: Ag⁺ + e⁻ → Ag (E°(Ag⁺/Ag) = 0 V) 2H⁺ + 2e⁻ → H₂ (E°(H⁺/H₂) = -0.80 V) The original cell potential (E°) can be calculated using the formula: E° = E°(products) - E°(reactants) = E°(Ag⁺/Ag) - E°(H⁺/H₂) = 0.80 V - 0 V = 0.80 V Calculating the new cell potential (E°'): E°' = E°'(Ag⁺/Ag) - E°'(H⁺/H₂) = 0 V - (-0.80 V) = 0.80 V Comparing the original cell potential (E°) with the new cell potential (E°'), we notice that they are the same (0.80 V).