Consider the following reduction processes : Al³⁺ + 3e⁻ → Al(s), E° = −1.66 V Fe³⁺ + e⁻ → Fe²⁺, E° = +0.77 V Co³⁺ + e⁻ → Co²⁺, E° = +1.81 V Cr³⁺ + 3e⁻ → Cr(s), E° = −0.74 V The tendency to act as reducing agent decreases in the order :

Reduction potentials and order of reducing strength of elements

Q. Consider the following reduction processes :

Al³⁺ + 3e⁻ → Al(s), E° = −1.66 V
Fe³⁺ + e⁻ → Fe²⁺, E° = +0.77 V
Co³⁺ + e⁻ → Co²⁺, E° = +1.81 V
Cr³⁺ + 3e⁻ → Cr(s), E° = −0.74 V

The tendency to act as reducing agent decreases in the order :

A. Al > Fe²⁺ > Cr > Co²⁺

B. Al > Cr > Co²⁺ > Fe²⁺

C. Cr > Fe²⁺ > Al > Co²⁺

D. Al > Cr > Fe²⁺ > Co²⁺

Correct Answer: Al > Cr > Fe²⁺ > Co²⁺

Explanation

A reducing agent is a species that loses electrons easily. Hence, the tendency to act as a reducing agent depends on how easily a substance gets oxidised.

Standard reduction potential (E°) gives the tendency of a species to get reduced. Therefore:

More negative the reduction potential → more easily it gets oxidised → stronger reducing agent.

Now, list the given standard reduction potentials:

$$ \begin{aligned} \text{Al}^{3+}/\text{Al} & : E^\circ = -1.66\ \text{V} \\ \text{Cr}^{3+}/\text{Cr} & : E^\circ = -0.74\ \text{V} \\ \text{Fe}^{3+}/\text{Fe}^{2+} & : E^\circ = +0.77\ \text{V} \\ \text{Co}^{3+}/\text{Co}^{2+} & : E^\circ = +1.81\ \text{V} \end{aligned} $$

Among these, aluminium has the most negative reduction potential, so it is the strongest reducing agent.

Chromium has the next lower (negative) value, so it comes next.

Between Fe²⁺ and Co²⁺, Fe²⁺ has a lower reduction potential than Co²⁺, so Fe²⁺ is a stronger reducing agent than Co²⁺.

Hence, the decreasing order of reducing strength is:

Al > Cr > Fe²⁺ > Co²⁺

Explanation

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