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 substance most reduced?
In the given reactions, the most reduced substance is the one that gains electrons and goes from a higher oxidation state to a lower oxidation state.
Looking at the reactions:
1) L^2+(aq) + 2K(s) = L(s) + 2K^+(aq)
In this reaction, L^2+ undergoes reduction and gets reduced to L(s), going from an oxidation state of 2+ to 0.
2) K(s) +N^+(aq) = no change
In this reaction, there is no change in oxidation state observed for any of the substances.
3) K+(aq) + K(s) = K(s) +M^+(aq)
In this reaction, K+ does not undergo any change in oxidation state.
4) N(s) + M^+(aq) = N^+(aq) + M(s)
In this reaction, N(s) undergoes oxidation and gets oxidized to N^+, going from an oxidation state of 0 to 1+.
From the given reactions, it is observed that the substance L goes from an oxidation state of 2+ to 0, indicating that it is the most reduced substance among L, K, M, and N.