The reduction potential values are given below :
${A l}^{3 +} / A l = - 1.67 v o l t, {M g}^{2 +} / M g = - 2.34 v o l t {C u}^{2 +} / C u = + 0.34 v o l t$
$I_{2} / {2 l}^{-} = + .053 v o l t$ . Which one is the best reducing agent ?

A l

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M g

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C u

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I_{2}

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Solution

The correct option is D $M g$
The metal which has the most negative reduction potential will lie to the top in the electrochemical series and will be a best reducing agent. Here, $M g$ has the most negative reduction potential hence it is the best reducing agent.
So, option B is correct.

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Similar questions

A l^{3 +} / A l = - 1.67 v o l t, M g^{2 +} / M g = - 2.34 v o l t

C u^{2 +} / C u = + 0.34 v o l t, I_{2} / 2 I^{-} = + 0.53 v o l t

M g^{2 +} / M g

- 2.37 V

C u^{2 +} / C u

0.337

E_{c e l l}^{o}

M g + C u^{2 +} \to M g^{2 +} + C u

The oxidation and reduction half reactions of the reaction

2Al_(s) + 3Cu²⁺_(aq) → 2Al³⁺_(aq) + 3Cu_(s) are

M g (s) + c u^{2 +} (a q) ⇌ C u (s) + M g^{2 +} (a q)

M g

C u

- 2.37

+ 0.34 V

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