The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is Fe ( aq ) 2 + + 2 e − → Fe ( s ) , E ∘ = − 0.44 V Cr 2 O 7 2 − ( aq

Question

The correct value of cell potential in volt for the reaction that occurs when the following two half cells are connected, is $Fe_{(aq)}^{2 +} + 2 e^{-} \to Fe (s), E^{\circ} = - 0.44 V Cr_{2} O_{7}^{2 -} (aq) + 14 H^{+} + 6 e^{-} \to 2 Cr^{3 +} + 7 H_{2} O, E^{0} = + 1.33 V$ NEET 2023 (Manipur)

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Answer:

C

Solution:

$E_{cell}^{0} = E_{cathode} - E_{anode} = 1.33 - (- 0.44) = 1.33 + 0.44 = + 1.77 V$

TestHub · Accepted Answer

E cell 0 ​ ​ = E cathode ​ − E anode ​ = 1.33 − ( − 0.44 ) = 1.33 + 0.44 = + 1.77 V ​

Chemistry - Electrochemistry Question with Solution | TestHub

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