Using the data given below find out the strongest reducing agent.

${E^{o}}_{{C r}_{2} O_{4} / {C r}^{3 +}} = 1.33 V$ , ${E^{o}}_{{C l}_{2} / {C l}^{-}} = 1.36 V$ ,
${E^{o}}_{M n O_{4}^{-} / {M n}^{2 +}} = 1.51 V$ , ${E^{o}}_{{C r}^{3 +} / C r} = - 0.74 V$

{C l}^{-}

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{M n}^{2 +}

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C r

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{C r}^{3 +}

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Solution

The correct option is A $C r$
The strongest reducing agent is one which is oxidised most easily, i.e., which has the highest oxidation potential.

$C r / {C r}^{3 +}$ has the highest oxidation potential, i.e., ( $+ 0.74$ ) so, it is oxidised most easily and, therefore, it is the strongest reducing agent.

Hence, option C is correct.

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Similar questions

E_{C l_{2} / C l^{-}}^{o} = 1.36 V, E_{C r^{3 +} / C r}^{\circ} = - 0.74 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{o} = 1.33 V, E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 V

E_{C r^{3 +} / C r}^{\circ} = - 0.74 V; E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = + 1.51 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{\circ} = + 1.33 V; E_{C l / C l^{-}}^{\circ} = + 1.36 V

E_{{C l}_{2} / {C l}^{-}} = 1.36 V

E_{{C r}^{3 +} / C r} = - 0.74 V

E_{{C r}_{2} O_{7}^{2 -} / {C r}^{3 +}} = 1.33 V

E_{{M n O_{4}}^{-} / {M n}^{2 +}} = 1.51 V

E_{C l_{2} / C l^{-}}^{\circ} = 1.36 V, E_{C r^{3 +} / C r} = - 0.74 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}} = 1.33 V, E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = 1.51 V

E_{C l_{2} / C l^{-}}^{\circ} = 1.36 V, E_{C r^{3 +} / C r}^{\circ} = - 0.74 V

E_{C r_{2} / O_{7}^{2 -} / C r^{3 +}}^{\circ} = 1.33 V, E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = 1.51 V

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