Electrochemistry — JEE Main practice

25 questions

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Sample questions with solutions

Q1 · 2026

For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature?

Answer: D

In a Daniell cell operating in a closed circuit, properties like electrode concentrations and the instantaneous cell potential ( $E_{\text{cell}}$ ) change as the reaction proceeds and current flows, since the Nernst equation depends on the changing reactant/product concentrations.
However, the standard cell potential ( $E^\circ_{\text{cell}}$ ) is defined for standard conditions (1 M concentrations, standard temperature) and is a fixed thermodynamic quantity — it does not change as time passes, since it doesn't depend on the actual instantaneous concentrations of the cell.
Therefore, the correct plot must show $E^\circ_{\text{cell}}$ remaining constant over time, while any instantaneous or concentration-dependent property may vary.

Hence, the answer is Option D.

Q2 · 2026

Consider the following reduction processes :

$\begin{aligned} & \mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}(\mathrm{~s}), \mathrm{E}^0=-1.66 \mathrm{~V} \\ & \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}, \mathrm{E}^0=+0.77 \mathrm{~V} \\ & \mathrm{Co}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Co}^{2+}, \mathrm{E}^0=+1.81 \mathrm{~V} \\ & \mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Cr}(\mathrm{~s}), \mathrm{E}^0=-0.74 \mathrm{~V} \end{aligned}$

The tendency to act as reducing agent decreases in the order :

A.
$\mathrm{Al}>\mathrm{Fe}^{2+}>\mathrm{Cr}>\mathrm{Co}^{2+}$
B.
$\mathrm{Al}>\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Co}^{2+}$
C.
$\mathrm{Al}>\mathrm{Cr}>\mathrm{Co}^{2+}>\mathrm{Fe}^{2+}$
D.
$\mathrm{Cr}>\mathrm{Fe}^{2+}>\mathrm{Al}>\mathrm{Co}^{2+}$

Answer: B

A reducing agent is the species that gets oxidized (donates electrons). A more negative reduction potential for a couple means the reduced form is a stronger reducing agent (it's more easily oxidized), equivalently, a higher oxidation potential means a stronger reducing agent.
Compute the oxidation potentials (reverse of each given reduction, with sign flipped):

\text{Al}(s)\rightarrow\text{Al}^{3+}+3e^-,\quad E^\circ_{\text{ox}}=+1.66\text{ V}

\text{Cr}(s)\rightarrow\text{Cr}^{3+}+3e^-,\quad E^\circ_{\text{ox}}=+0.74\text{ V}

\text{Fe}^{2+}\rightarrow\text{Fe}^{3+}+e^-,\quad E^\circ_{\text{ox}}=-0.77\text{ V}

\text{Co}^{2+}\rightarrow\text{Co}^{3+}+e^-,\quad E^\circ_{\text{ox}}=-1.81\text{ V}

Arranging these in decreasing order of oxidation potential (i.e., decreasing reducing strength):

+1.66>+0.74>-0.77>-1.81

So the reducing strength decreases as:

\text{Al}>\text{Cr}>\text{Fe}^{2+}>\text{Co}^{2+}

Hence, the answer is Option B.

Q3 · 2026

In the given electrochemical cell, $\mathrm{Ag}(\mathrm{s})|\mathrm{AgCl}(\mathrm{s})| \mathrm{FeCl}_2(\mathrm{aq}), \mathrm{FeCl}_3(\mathrm{aq}) \mid \mathrm{Pt}(\mathrm{s})$ at 298 K , the cell potential ( $\mathrm{E}_{\text {cell }}$ ) will increase when :

A. Concentration of $\mathrm{Fe}^{2+}$ is increased.

B. Concentration of $\mathrm{Fe}^{3+}$ is decreased.

C. Concentration of $\mathrm{Fe}^{2+}$ is decreased.

D. Concentration of $\mathrm{Fe}^{3+}$ is increased.

E. Concentration of $\mathrm{Cl}^{-}$ is increased.

Choose the correct answer from the options given below :

A.
A and B Only
B.
B Only
C.
C, D and E Only
D.
A and E Only

Answer: C

The overall cell reaction for this setup is:

\text{Cl}^-(aq)+\text{Ag}(s)+\text{Fe}^{3+}(aq)\rightarrow\text{Fe}^{2+}(aq)+\text{AgCl}(s)

Applying the Nernst equation:

E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.059}{1}\log\frac{[\text{Fe}^{2+}]}{[\text{Cl}^-][\text{Fe}^{3+}]}

Since $E_{\text{cell}}$ increases as the argument of the logarithm decreases, we need $\frac{[\text{Fe}^{2+}]}{[\text{Cl}^-][\text{Fe}^{3+}]}$ to decrease.
This happens when $[\text{Fe}^{2+}]$ decreases (matches C), when $[\text{Cl}^-]$ increases (matches E), or when $[\text{Fe}^{3+}]$ increases (matches D).
Conversely, increasing $[\text{Fe}^{2+}]$ (A) or decreasing $[\text{Fe}^{3+}]$ (B) would increase the log argument and decrease $E_{\text{cell}}$ , so these do not qualify.

Hence, the correct combination is C, D and E only — Option C.

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