Question

Suppose \(E_{\text {red }}^{\circ}\) for \(\mathrm{H}^{+} \longrightarrow \mathrm{H}_{2}\) were taken to be \(0.300 \mathrm{~V}\) instead of \(0.000 \mathrm{~V}\). What would be (a) \(E_{\mathrm{ox}}^{\circ}\) for \(\mathrm{H}_{2} \longrightarrow \mathrm{H}^{+} ?\) (b) \(E_{\text {red }}^{\circ}\) for \(\mathrm{Br}_{2} \longrightarrow \mathrm{Br}^{-}\) ? (c) \(E^{\circ}\) for the cell in \(34(\mathrm{c})\) ? Compare your answer with that obtained in \(34(\mathrm{c})\).

Short answer

Answer: The standard oxidation potential for the reverse reaction H₂ → H⁺ is -0.300 V, and the new standard reduction potential for Br₂ → Br⁻ is 1.387 V.

Step-by-step solution

(a) Standard oxidation potential for H₂ → H⁺

To find \(E_{\mathrm{ox}}^{\circ}\) for the oxidation half-reaction, we will use the relation between standard reduction potential and standard oxidation potential: \(E_{\mathrm{ox}}^{\circ}=-E_{\text {red }}^{\circ}\). As we are given \(E_{\text {red }}^{\circ}\) for H⁺ → H₂, we can calculate the \(E_{\mathrm{ox}}^{\circ}\) for the reverse reaction H₂ → H⁺: \(E_{\mathrm{ox}}^{\circ}=-0.300 \mathrm{~V}\). So, the standard oxidation potential for H₂ → H⁺ is -0.300 V.

(b) Standard reduction potential for Br₂ → Br⁻

The new standard reduction potential for Br₂ → Br⁻ can be found as follows: Since standard reduction potentials are arbitrary, we can only find the difference between two potentials, not the absolute values. But we know that under the old system with \(E_{\mathrm{H}^{+}/\mathrm{H}_{2}}^{\circ}=0.000 \mathrm{~V}\), the standard reduction potential for Br₂ → Br⁻ was \(E_{\mathrm{Br}_{2}/\mathrm{Br}^{-}}^{\circ, \text{old}}=1.087 \mathrm{~V}\). Thus, we can find the change in standard reduction potential based on the change in H⁺ → H₂: \(\Delta E_{\mathrm{Br}_{2}/\mathrm{Br}^{-}}=E_{\mathrm{H}^{+}/\mathrm{H}_{2}}^{\circ, \text{new}} - E_{\mathrm{H}^{+}/\mathrm{H}_{2}}^{\circ, \text{old}}=0.300 \mathrm{~V} - 0.000 \mathrm{~V}\) Now, we can calculate the new standard reduction potential for Br₂ → Br⁻: \(E_{\mathrm{Br}_{2}/\mathrm{Br}^{-}}^{\circ, \text{new}}=E_{\mathrm{Br}_{2}/\mathrm{Br}^{-}}^{\circ, \text{old}}+\Delta E_{\mathrm{Br}_{2}/\mathrm{Br}^{-}}=1.087 \mathrm{~V} + 0.300 \mathrm{~V}\) So, the new standard reduction potential for Br₂ → Br⁻ is 1.387 V.

(c) New standard cell potential for the cell in 34(c)

We do not have enough information about the cell in 34(c) to calculate the new standard cell potential (\(E^{\circ}\)) for that cell.