Answer :

(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.


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