Metals L, K, M and N and their are selectively reacted and yield the following data:
L^2+(aq) + 2K(s) = L(s) + 2K^+(aq)
K(s) +N^+(aq) = no change
K+(aq) + K(s) = K(s) +M^+(aq)
N(s) + M^+(aq) = N^+(aq) + M(s)
What is strongest oxiding agent?
To determine the strongest oxidizing agent among metals L, K, M, and N, we need to compare their tendency to lose electrons. The stronger a metal is at losing electrons, the stronger of an oxidizing agent it will be.
Looking at the given reactions, we can analyze each metal's behavior:
1. L^2+(aq) + 2K(s) = L(s) + 2K^+(aq)
In this reaction, metal L is reduced from its 2+ oxidation state to its elemental form (L(s)). Metal K is oxidized from its elemental form to its 1+ oxidation state. This indicates that metal L is a weaker reducing agent than metal K. However, we are interested in identifying the strongest oxidizing agent, so this reaction does not provide information about metal L's oxidizing potential.
2. K(s) + N^+(aq) = no change
In this reaction, metal K remains unchanged. It neither gains nor loses electrons. Therefore, this reaction does not provide information about metal K's oxidizing potential.
3. K+(aq) + K(s) = K(s) + M^+(aq)
In this reaction, metal K is reduced from its 1+ oxidation state to its elemental form (K(s)). Metal M is oxidized from its elemental form to its 1+ oxidation state. This indicates that metal K is a weaker reducing agent than metal M. Again, we are interested in identifying the strongest oxidizing agent, so this reaction does not provide information about metal K's oxidizing potential.
4. N(s) + M^+(aq) = N^+(aq) + M(s)
In this reaction, metal N is oxidized from its elemental form to its 1+ oxidation state. Metal M is reduced from its 1+ oxidation state to its elemental form (M(s)). This indicates that metal N is a stronger reducing agent than metal M. Hence, metal N will be the strongest oxidizing agent among the given metals.
Therefore, metal N is the strongest oxidizing agent.