Question

Consider the following half-reactions: ε=-0.315 NAD⁺ + H⁺ + 2e --> NADH ε = 0.815 1/2 O₂ + 2H⁺ + 2e --> H₂O What is the Δ8" for the following redox equation?

Answer 1

The standard redox potential (ΔE°) for the reaction between NO₃⁻. and Cu is calculated as 1.30 V by subtracting the reversed standard electrode potential of the anode Cu/Cu²⁺ (-0.34 V) from that of the cathode NO₃⁻/NO (+0.96 V).

Step-by-step explanation:

The standard redox potential (ΔE°) for a redox reaction can be calculated by subtracting the standard electrode potential of the anode (oxidation half-reaction) from the standard electrode potential of the cathode (reduction half-reaction). To calculate ΔE°, we use the given half-reaction potentials:

• NO₃⁻. + 4H⁺ + 3e¯ → NO, with a standard electrode potential (E1/2) of 0.96 V.
• Cu → Cu²⁺ + 2 e¯, with E1/2 reversed to -0.34 V because the reaction is in the oxidation direction in the overall redox process.

To find the ΔE°, we apply the formula:

ΔE° = E°(cathode) - E°(anode)

ΔE° = 0.96 V - (-0.34 V) = 1.30 V

Therefore, the standard redox potential for the reaction between NO₃⁻ and Cu is 1.30 V.

Answer 2

The standard cell potential (E0) for the given redox reaction is 0.62 V.

Step-by-step explanation:

The given redox reaction consists of two half-reactions:

Oxidation half-reaction: Cu → Cu2+ + 2e-

Reduction half-reaction: NO3 + 4H+ + 3e- → NO + 2H2O

To calculate the standard cell potential (E0), we need to compare the reduction potentials of each half-reaction. The reduction potential for Cu → Cu2+ + 2e- is 0.34 V, and for NO3 + 4H+ + 3e- → NO + 2H2O is 0.96 V.

The standard cell potential (E0) can be calculated by adding the reduction potentials of the reduction and oxidation half-reactions:

E0 = E0(reduction) - E0(oxidation) = 0.96 V - 0.34 V = 0.62 V