Chemistry · Electrochemistry

JEE Main 2024 — 9 April, Shift 1 — Question 78

The standard reduction potentials at 298 K for the following half cells are given below :

Cr2O72−+14H++6e−→2Cr3++7H2O,E∘=1.33 V\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.33 \mathrm{~V}

Fe3+(aq)+3e−→FeE∘=−0.04 V\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathrm{E}^{\circ}=-0.04 \mathrm{~V}

Ni2+(aq)+2e−→NiE∘=−0.25 V\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} \quad \mathrm{E}^{\circ}=-0.25 \mathrm{~V}

Ag+(aq)+e−→AgE∘=0.80 V\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \quad \mathrm{E}^{\circ}=0.80 \mathrm{~V}

Au3+(aq)+3e−→AuE∘=1.40 V\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \quad \mathrm{E}^{\circ}=1.40 \mathrm{~V}

Consider the given electrochemical reactions, The number of metal(s) which will be oxidized be Cr2O72−\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}, in aqueous

solution is \qquad

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

Fe, Ni, Ag will be oxidized due to lower S.R.P.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Chemistry
Chapter
Electrochemistry
Topic
Nernst Equation and Electrochemical Series
The standard reduction potentials at 298 K for the following half… | JEE Main 2024 PYQ with Solution · DhiX AI