The standard redox potentials E0 of the following system are I-MnO 4-8 H -5 e - Mn 2-4 H 2 OE 0-1-5 VII- Sn 2- S n 4-2 e - E 0-0-15 VIII- Cr 2 O 72-14 H -6 e - 2 Cr 2-7 H 2 O E 0-1-33 VIV - Ce 3- Ce 4- e - E 0-1-61 VThe oxidizing power of the various species are related asA- Cr 2 O 2- 7- MnO 4-B- Ce 4- Sn 4-C- Ce 4- M nO 4-4D- MnO 4- Sn 4-

The correct options are B Ce^4+ gt;Sn^4+ C Ce^4+ gt;MnO_4^ D MnO^ _4 gt;Sn^4+Electrode with highest +SRP will be strongest oxidant ...

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Application of Electrolysis

The standard ...

Question

The standard redox potentials $E^{0}$ of the following system are
$I . M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{0} = 1.5 V$
$I I . S n^{2 +} \to S n^{4 +} + 2 e^{-} E^{0} = - 0.15 V$
$I I I . C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{2 +} + 7 H_{2} O E^{0} = 1.33 V$
$I V . C e^{3 +} \to C e^{4 +} + e^{-} E^{0} = - 1.61 V$
The oxidizing power of the various species are related as

C r_{2} O^{2 -} 7 > M n O_{4}^{-}

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C e^{4 +} > S n^{4 +}

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C e^{4 +} > M n O_{4}^{-}

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M n O_{4}^{-} > S n^{4 +}

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Solution

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E^{0}

I . M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O E^{0} = 1.5 V

I I . S n^{2 +} \to S n^{4 +} + 2 e^{-} E^{0} = - 0.15 V

I I I . C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{2 +} + 7 H_{2} O E^{0} = 1.33 V

I V . C e^{3 +} \to C e^{4 +} + e^{-} E^{0} = - 1.61 V

^{o}

E^{0}