NEET2026ChemistryChapterActual
Given below are half cell reactions : Cr₂O₇²⁻ + 14H^+ + 6e^- 2Cr³⁺ + 7H₂O , E^ _ Cr³⁺/Cr₂O₇²⁻ = -1.33 V Cl₂ + 2e^- 2Cl^- , E^ _ Cl₂/Cl^- = +1.36 V Will the dichromate ion, Cr₂O₇²⁻ , oxidize Cl^- to Cl₂ gas in an acidic medium under standard conditions?
Options
- ANo, because E^ _ cell = -0.03 V
- BYes, because E^ _ cell = +2.69 V
- CNo, because E^ _ cell = -2.69 V
- DYes, because E^ _ cell = +0.03 V
Correct answer
A. No, because E^ _ cell = -0.03 V
Step-by-step solution
The given standard oxidation potential of Cr³⁺ to Cr₂O₇²⁻ is E^ _ Cr³⁺/Cr₂O₇²⁻ = -1.33 V . The standard reduction potential of Cr₂O₇²⁻ is E^ _ Cr₂O₇²⁻/Cr³⁺ = +1.33 V . The standard reduction potential of Cl₂ is E^ _ Cl₂/Cl^- = +1.36 V . For the oxidation of Cl^- by Cr₂O₇²⁻ , the dichromate ion undergoes reduction (cathode) and the chloride ion undergoes oxidation (anode). The standard cell potential is calculated as: E^ _ cell = E^ _ cathode - E^ _ anode E^ _ cell = E^ _ Cr₂O₇²⁻/Cr³⁺ - E^ _ Cl₂/Cl^- E^ _ cell = 1.3