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E⁰ for two reactions are given below: ${Cr}^{+ 3} + 3 e^{-} \to Cr, E^{0} = - 0.74 V; {OCl}^{-} + H_{2} O + 2 e^{-} \to {Cl}^{-} + 2 {OH}^{- 1}$ ,
E⁰ = 0.94V. What will be the E⁰ for $3 {OCl}^{-} + 2 Cr + 3 H_{2} O \to 2 {Cr}^{+ 3} + 3 {Cl}^{-} + 6 {OH}^{-}$

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An Intiative by Sri Chaitanya

–1.68 V

–0.20V

1.68 V

+0.20 V

answer is B.

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Detailed Solution

$E_{Cell} = E_{cathode}^{0} - E_{anode}^{0}$

$= 0.94 - (- 0.74) = 1.68 V$

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