(i) F₂ → (c) non-metal which is the best oxidizing agent

(ii) Li → (a) metal is the strongest reducing agent

(iii) Au³⁺ → (g) metal ion which is an oxidizing agent

(iv) Br^– → (e) anion that can be oxidized by Au³⁺

(v) Au → (d) unreactive metal

(vi) Li ⁺→ (b) metal ion which is the weakest oxidizing agent

(vii) F^– → (f) anion which is the weakest reducing Agent

(i) We know, Fluoride is a non-metal and the standard electrode potential is +2.87V. High positive value means they are good oxidizing agent. Thus, F₂ get easily reduced to F^-.

(ii) Li⁺/Li have a standard electrode potential of -3.5V. Thus, this radox couple is a stronger reducing agent than the H⁺/H₂ radox couple. In other words, they have the most negative potential value among the above chemical elements which makes Li metal a strongest reducing agent.

(iii) The standard electrode potential for Au³⁺/Au is + 1.4V. Thus, they are a weaker reducing agent than the H⁺/H₂ radox couple. Positive standard electrode potential makes Au³⁺ a good oxidizing agent.

(iv) The standard electrode potential for this radox couple is +1.09V. Clearly, one can say that the standard electrode potential of this radox couple is less than that of Au³⁺/Au. Thus, Br^- is weak oxidizing agent and stronger reducing agent than Au³⁺. Therefore, Br^- anion can be oxidized by Au³⁺.

(v) Au³⁺ + 3e^-→ Au_(S)

Since, Au is in the solid state. Thus, its standard electrode potential will be zero. So, Au is an unreactive metal.

(vi) Li⁺/Li have a standard electrode potential of -3.5V. Thus, this radox couple is a stronger reducing agent than the H⁺/H₂ radox couple. In other words, Li⁺ metal ion is the weakest oxidizing agent.

(vii) F₂/F^- have the standard electrode potential equal to +2.87V. High positive value of standard radox potential indicates that F₂ is the best oxidizing agent, whereas, F^- is the weakest reducing Agent.