Question

For each of these overall cell reactions, write the oxidation and reduction half-reactions, calculate the standard cell potential (E∘), and determine if the reaction is spontaneous or not.

Fe₃⁺(aq)+CO₂⁺(aq)→Fe₂⁺(aq)+Co₃2⁻​(aq)

Oxidation half-reaction: Reduction half-reaction: E∘ =
Cell spontaneous?

Answer 1

To determine oxidation and reduction half-reactions and whether a reaction is spontaneous, one must know the oxidation states of reactants and products,

Step-by-step explanation:

The initial reaction provided seems to contain a typo with the ion 'CO₂⁺⁺' and 'Co₃₂⁻' which are not common ions. Assuming that it might be 'Co²⁺' for cobalt(II) and 'Co³⁺' for cobalt(III), let's correct it to:

Fe³⁺(aq) + Co²⁺(aq) → Fe²⁺(aq) + Co³⁺(aq)

The oxidation and reduction half-reactions will be identified based on the change in oxidation states of iron and cobalt. The standard cell potential (E°) is calculated using the standard reduction potentials from a reference table (which we assume to be provided).

Oxidation half-reaction:
Fe²⁺(aq) → Fe³⁺(aq) + e⁻

Reduction half-reaction:
Co³⁺(aq) + e⁻ → Co²⁺(aq)

To calculate the E° cell, we would use the standard electrode potentials for Fe³⁺/Fe²⁺ and Co³⁺/Co²⁺ couples from a reference table, adding them if the tabulated values are for the reduction half-reactions or subtracting if one is reversed (oxidation).

A positive value of E° cell would indicate that the reaction is spontaneous under standard conditions.