Question

Match the following:

List-I (Transition Metal)		List-II \(\left(E^\circ_{\mathrm{M^{2+}/M}} \text{ in V}\right)\)
A	\(\mathrm{Cr}\)	I	\(-1.18\)
B	\(\mathrm{Co}\)	II	\(-0.28\)
C	\(\mathrm{Mn}\)	III	\(-0.44\)
D	\(\mathrm{Fe}\)	IV	\(-0.90\)

The correct answer is

• A. A--IV, B--I, C--II, D--III
• B. A--I, B--II, C--IV, D--III
• C. A--III, B--IV, C--II, D--I
• D. A--IV, B--II, C--I, D--III

Hint:

The important standard reduction potentials are \[ \boxed{ \begin{aligned} \mathrm{Mn^{2+}/Mn}&=-1.18\ \text{V}, \mathrm{Cr^{2+}/Cr}&=-0.90\ \text{V}, \mathrm{Fe^{2+}/Fe}&=-0.44\ \text{V}, \mathrm{Co^{2+}/Co}&=-0.28\ \text{V}. \end{aligned} } \] These values are frequently asked in competitive examinations.

Answer 1

The Correct Option is D

Step 1: Recall the standard reduction potentials.& nbsp;

\[ \begin{aligned} \mathrm{Mn^{2+}/Mn} & amp;=-1.18\ \text{V},\\ \mathrm{Cr^{2+}/Cr} & amp;=-0.90\ \text{V},\\ \mathrm{Fe^{2+}/Fe} & amp;=-0.44\ \text{V},\\ \mathrm{Co^{2+}/Co} & amp;=-0.28\ \text{V}. \end{aligned} \]

Step 2: Match the values.

Hence,

\[ \begin{aligned} A & amp;\rightarrow IV,\\ B & amp;\rightarrow II,\\ C & amp;\rightarrow I,\\ D & amp;\rightarrow III. \end{aligned} \]

Thus, the correct matching is

\[ \boxed{\text{A--IV,\; B--II,\; C--I,\; D--III}.} \]

Therefore, the correct option is

\[ \boxed{(D)}. \]