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

{M n}^{2 +}

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{C r}^{3 +}

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{C l}^{-}

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C r

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Solution

The correct option is C $C r$
An oxidizing agent can be defined as a substance which gains electrons and is reduced in a chemical reaction and a reducing agent is the vice versa.
In terms of potential the more negative the potential the more strong reducing agent and the more positive the potential the stronger oxidizing agent it will be.
In the question the most negative potential is of $C r^{3 +} = - 0.74 V$
Therefore $C r$ is the strongest reducing agent.

Correct option is B .

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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 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

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^{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

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

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