Question

1. Br₂(s) + 2K(s) → 2Br⁻(aq) + 2K⁺(aq) ; SRP Br₂(s) > SRP K⁺(aq)

2. Hg²⁺(aq) + 2K(s) → Hg(l) + 2K⁺ ; SRP Hg²⁺(aq) > SRP K⁺(aq)

3. NO REACTION (therefore reverse reaction is spontaneous: Co²⁺(aq) + 2K(s) → Co(s) + 2K⁺(aq)) ; SRP Co²⁺(aq) > SRP K⁺(aq)

4. NO REACTION (therefore reverse reaction is spontaneous: Hg²⁺(aq) + Co(s) → Hg(l) + Co²⁺(aq)) ; SRP Hg²⁺(aq) > SRP Co²⁺(aq)

5. Br₂(s) + Hg(l) → 2Br⁻(aq) + Hg²⁺(aq) ; SRP Br₂(s) > SRP Hg²⁺(aq)

therefore the overall order is:
E°(Br₂) > E°(Hg²⁺) > E°(Co²⁺) > E°(K⁺)

Answer 1

The question deals with determining the order of oxidizing strength among chemical species based on standard reduction potentials (SRP), which is a key concept in electrochemistry.

Step-by-step explanation:

The question provided involves a series of chemical reactions and standard reduction potentials (SRP) to determine the order of oxidizing strength of several chemical species. This is a fundamental concept in electrochemistry, which is a branch of chemistry that deals with the relationships between electricity and chemical reactions, especially oxidation-reduction (redox) reactions.

Through the given reactions and SRP values, an overall order of oxidizing agent strength is established as E°(Br2) > E°(Hg2+) > E°(Co2+) > E°(K+), based on which species will spontaneously oxidize potassium (K).

A higher SRP indicates a greater tendency to accept electrons and thus a stronger oxidizing agent. The provided reactions illustrate how these properties are applied to predict the direction and spontaneity of redox reactions.