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Given, $E_{C l_{2} / C l^{-}}^{0} = 1.36 V, E_{C r^{3 +} / C r}^{0} = - 0.74 V, E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{0} = 1.33 V, E_{M n O_{4}^{-} / M n^{2 +}}^{0} = 1.51 V$ Among the following, the strongest reducing agent is

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$M n^{2 +}$

$C r^{3 +}$

$C l^{-}$

answer is A.

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Detailed Solution

$\begin{array}{l} E_{M n^{2 +} / M n O_{4}^{-}}^{0} = - 1.51 V \\ E_{C l^{-} / C l_{2}}^{0} = - 1.36 V; E_{C r^{3 +} / C_{2} O_{7}^{2 -}}^{0} = - 1.33 V \\ E_{C r / C r^{3 +}}^{0} = + 0.74 V \end{array}$

Since Cr is having highest oxidation potential, Cr is the best reducing agent.

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