COMEDK 2025 Morning Shift | Electrochemistry Question 1 | Chemistry

What is the reduction potential of a half-cell consisting of a Pt electrode dipped in $2.2 \mathrm{M} \mathrm{Fe}^{2+}$ and $0.04 \mathrm{M} \mathrm{Fe}^{3+}$ solution where the reaction taking place is conversion of $\mathrm{Fe}^{3+}$ ions to $\mathrm{Fe}^{2+}$ ?

$$\mathrm{E}^0\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.771 \mathrm{~V}$$

0.598 V

0.723 V

0.668 V

0.719 V

COMEDK 2025 Morning Shift

MCQ (Single Correct Answer)

-0

A current of 2.5 amperes is passed through 800 ml of 0.48 M solution of $\mathrm{CuSO}_4$ for 1.0 hour with a current efficiency of $80 \%$. If the volume of the solution remains unchanged, what is the final molarity of the solution?

0.386

0.315

0.433

0.298

COMEDK 2025 Morning Shift

MCQ (Single Correct Answer)

-0

The EMF of the cell $\mathrm{Al} / \mathrm{Al}^{3+}(0.01 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.02 \mathrm{M}) / \mathrm{Fe}$ is 1.209 V . The EMF of the cell can be increased by

increasing the concentration of $\mathrm{Al}^{3+}$ and $\mathrm{Fe}^{2+}$

increasing the concentration of $\mathrm{Al}^{3+}$

increasing the concentration of $\mathrm{Fe}^{2+}$

decreasing the concentration of $\mathrm{Al}^{3+}$ and $\mathrm{Fe}^{2+}$

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

Identify the correct statement regarding corrosion of iron rod left exposed to atmosphere.

The reaction occurring at the cathodic area is:

$$ \mathrm{O}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})+4 \mathrm{e} \rightarrow 4 \mathrm{OH}^{-} $$

The reaction occurring at the cathodic area is:

$$ \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e} \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) ; \mathrm{E}^0=+1.23 \mathrm{~V} $$

Reaction occurring at anodic area is:

$$ 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+}+4 \mathrm{e} $$

The overall reaction occurring during the corrosion process is:

$$ 2 \mathrm{Fe}_{(\mathrm{S})}+3 / 2 \mathrm{O}_2(\mathrm{~g})+6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Fe}^{3+}+3 \mathrm{H}_2 \mathrm{O} ; \mathrm{E}^0=-1.23 \mathrm{~V} $$

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