NEET 2023 Phase 2 · ChemistryGalvanic & EMFPrevious Year Question

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

Answer: (C) $+ 1.77 V$ . \textbf{Answer:} (C) $+ 1.77 V$ \textbf{Solution:} The couple with the higher reduction potential acts as the cathode (reduction), and the other as the anode (oxidation).

A. $+ 0.01 V$
B. $+ 0.89 V$
C. $+ 1.77 V$ ✓
D. $+ 2.65 V$

Correct Answer

Solution & Explanation

\textbf{Answer:} (C) $+ 1.77 V$ \textbf{Solution:} The couple with the higher reduction potential acts as the cathode (reduction), and the other as the anode (oxidation). Cathode: $Cr X_{2} O X_{7} X^{2 -} / Cr X^{3 +}$ , $E^{\circ} = + 1.33 V$ Anode: $Fe X^{2 +} / Fe$ , $E^{\circ} = - 0.44 V$ $E_{ce l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{an o d e}^{\circ} = 1.33 - (- 0.44) = + 1.77 V$

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The correct value of cell potential (in volt) for the reaction that occurs when the following two half cells are connected is: FeX2+(aq)+2eX−​Fe(s), E∘=−0.44 V CrX2​OX7​X2−+14HX++6eX−​2CrX3++7HX2​O, E∘=+1.33 V

Solution & Explanation

More NEET PYQ solutions

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