Question
Question: Given that Eº values of Ag<sup>+</sup>/Ag, K<sup>+</sup>/ K, Mg<sup>2+</sup>/ Mg and Cr<sup>3+</sup>...
Given that Eº values of Ag+/Ag, K+/ K, Mg2+/ Mg and Cr3+/ Cr are 0.80V, –2.93V, –2.37V and –0.74V respectively. Therefore the order for the reducing power of the metal is –
A
Ag > Cr > Mg > K
B
Ag < Cr < Mg < K
C
Ag > Cr > K > Mg
D
Cr > Ag > Mg > K
Answer
Ag < Cr < Mg < K
Explanation
Solution
More the negative Eº value, larger the reducing power of the metal.