Given E-Cr3- Cr - -0-74 V- E-MnO4- Mn2- - - 1-51 V E-Cr2O72- Cr3- - - 1-33 V- E-Cl- Cl- - - 1-36 Vbased on the data given above- strongest oxidizing agent will be-

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Electrode Potential

Given E∘Cr3+...

Question

Given
$E_{C r^{3 +} / C r}^{\circ} = - 0.74 V; E_{M n O_{4}^{-} / M n^{2 +}}^{\circ} = + 1.51 V$
$E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{\circ} = + 1.33 V; E_{C l / C l^{-}}^{\circ} = + 1.36 V$
based on the data given above, strongest oxidizing agent will be__________.

M n^{2 +}

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M n O_{4}^{-}

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C l^{-}

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C r^{3 +}

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Solution

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E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{o} = 1.33

E_{C l_{2} / C l^{-}}^{o} = 1.36

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

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

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 r^{(} 3 +) / C r}^{0} = - 0.74 V; E_{M n O_{4}^{-} / M n^{2 +}}^{0} = 1.51 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{0} = 1.33 V; E_{C l / C l^{-}}^{0} = 1.36 V

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

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

E_{C r^{(} 3 +) / C r}^{0} = - 0.74 V; E_{M n O_{4}^{-} / M n^{2 +}}^{0} = 1.51 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{0} = 1.33 V; E_{C l / C l^{-}}^{0} = 1.36 V

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