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Friday, 9 October

JEE Main 2024 · ChemistryNumerical answerNumerical valueMediumApplication

JEE Main 9 April 2024, Shift 1, Chemistry Q85

Question 85 of 90 in this shift, Chemistry question 25 of 30, Section B.

The standard reduction potentials at 298 K for the following half cells are given below : Cr2O72−+14H++6e−→2Cr3++7H2O\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6e^- \rightarrow 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O}, E∘=1.33E^\circ = 1.33 V Fe3+(aq)+3e−→Fe\mathrm{Fe^{3+}(aq)} + 3e^- \rightarrow \mathrm{Fe} E∘=−0.04E^\circ = -0.04 V Ni2+(aq)+2e−→Ni\mathrm{Ni^{2+}(aq)} + 2e^- \rightarrow \mathrm{Ni} E∘=−0.25E^\circ = -0.25 V Ag+(aq)+e−→Ag\mathrm{Ag^+(aq)} + e^- \rightarrow \mathrm{Ag} E∘=0.80E^\circ = 0.80 V Au3+(aq)+3e−→Au\mathrm{Au^{3+}(aq)} + 3e^- \rightarrow \mathrm{Au} E∘=1.40E^\circ = 1.40 V Consider the given electrochemical reactions, The number of metal(s) which will be oxidized be Cr2O72−\mathrm{Cr_2O_7^{2-}}, in aqueous solution is ________.

Official answer

3

NTA final key.