The standard reduction potential (E°) values of \( Ag^+|Ag, K^+|K, Mg^{2+}|Mg \) and \( Cr^{3+}|Cr \) are 0.80 V, -2.93 V, -2.37 V and -0.74 V, respectively. The decreasing order of reducing power of the following metals is
(A) Ag
(B) K
(C) Mg
(D) Cr
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (C), (D), (A)
The standard reduction potential E° measures the tendency of the ion to gain electrons. A metal whose couple has a more negative E° loses electrons more easily, so it is a stronger reducing agent. Reducing power therefore decreases as E° becomes more positive.
(B) K: \( E^{\circ}(\ce{K+}|\ce{K}) = -2.93\ \mathrm{V} \), the most negative, so K is the strongest reducing agent: \( \ce{K -> K+ + e-} \).
(C) Mg: \( E^{\circ}(\ce{Mg^2+}|\ce{Mg}) = -2.37\ \mathrm{V} \).
(D) Cr: \( E^{\circ}(\ce{Cr^3+}|\ce{Cr}) = -0.74\ \mathrm{V} \).
(A) Ag: \( E^{\circ}(\ce{Ag+}|\ce{Ag}) = +0.80\ \mathrm{V} \), the most positive, so Ag is the weakest reducing agent; \( \ce{Ag+} \) is instead a good oxidising agent.
Shortcut: K has the most negative and Ag the most positive E°, so the order starts with (B) and ends with (A); only Option (3) does this.