Electrochemistry Previous Year Questions — JEE Main Chemistry
40 past-year Electrochemistry questions from JEE Main (Chemistry) — page 7.
Q72 (2026)
Consider the following electrochemical cell at 298K
$$Pt | HSnO_{2}^{-}(aq) | Sn(OH)_{6}^{2-}(aq) || Bi_{2}O_{3}(s) | Bi(s)$$
If the reaction quotient at a given time is $10^{6}$, then the cell EMF $(E_{\text{cell}})$ is $x \times 10^{-1}\text{ V}$ (Nearest integer).
Given the standard half-cell reduction potential as
$$E_{Bi_{2}O_{3}/Bi,OH^{-}}^{0} = -0.44\text{ V} \text{ and } E_{Sn(OH)_{6}^{2-}/HSnO_{2}^{-},OH^{-}}^{0} = -0.90\text{ V}$$
### Related Formula
$$E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{n} \log Q$$
(Using 0.06 for approximation if needed, as per standard practice unless specified).
### Core Logic
First, calculate $E^{\circ}_{cell}$:
$$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$
$$E^{\circ}_{cell} = (-0.44\text{ V}) - (-0.90\text{ V}) = +0.46\text{ V}$$
Identify the number of transferred electrons ($n$) for the overall reaction.
Anode reaction: $Sn^{2+} \rightarrow Sn^{4+} + 2e^{-}$ (Tin goes from +2 in $HSnO_2^-$ to +4 in $Sn(OH)_6^{2-}$).
Cathode reaction: $Bi^{3+} + 3e^{-} \rightarrow Bi$ (Bismuth goes from +3 in $Bi_2O_3$ to 0).
To balance the electrons, we multiply anode by 3 and cathode by 2, yielding $n = 6$ total electrons transferred.
### Step 1: Apply Nernst Equation
Applying the Nernst equation (approximating 0.0591 to 0.06 for simplicity):
$$E_{cell} = E^{\circ}_{cell} - \frac{0.06}{n} \log_{10} Q$$
$$E_{cell} = 0.46 - \frac{0.06}{6} \log_{10}(10^6)$$
$$E_{cell} = 0.46 - 0.01 \times 6$$
$$E_{cell} = 0.46 - 0.06 = 0.40\text{ V}$$
### Step 2: Format Output
The problem asks for the answer in the form $x \times 10^{-1}\text{ V}$.
$$0.40 = 4 \times 10^{-1}$$
So, $x = 4$.
### Pattern Recognition
Be extremely careful finding 'n'. You must balance the half-reactions. A +2 to +4 change (n=2) and a +3 to 0 change (n=3) resolves to a common multiple of 6.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q67 (2026)
Consider the following reduction processes:
$\text{Al}^{3+} + 3e^- \rightarrow \text{Al}(\text{s})$, $E^0 = -1.66\text{ V}$
$\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$, $E^0 = +0.77\text{ V}$
$\text{Co}^{3+} + e^- \rightarrow \text{Co}^{2+}$, $E^0 = +1.81\text{ V}$
$\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}(\text{s})$, $E^0 = -0.74\text{ V}$
The tendency to act as reducing agent decreases in the order:
- $\text{Al} > \text{Cr} > \text{Fe}^{2+} > \text{Co}^{2+}$
- $\text{Al} > \text{Fe}^{2+} > \text{Cr} > \text{Co}^{2+}$
- $\text{Al} > \text{Cr} > \text{Co}^{2+} > \text{Fe}^{2+}$
- $\text{Cr} > \text{Fe}^{2+} > \text{Al} > \text{Co}^{2+}$
### Related Formula
$$\text{Reducing Power} \propto \frac{1}{\text{Standard Reduction Potential } (E^0_{red})}$$
### Core Logic
Step 1: List standard reduction potential values:
- $E^0(\text{Al}^{3+}/\text{Al}) = -1.66\text{ V}$
- $E^0(\text{Cr}^{3+}/\text{Cr}) = -0.74\text{ V}$
- $E^0(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}$
- $E^0(\text{Co}^{3+}/\text{Co}^{2+}) = +1.81\text{ V}$
Step 2: Arrange species in increasing order of $E^0_{red}$ (decreasing reducing power):
$$\text{Al} > \text{Cr} > \text{Fe}^{2+} > \text{Co}^{2+}$$
### Pattern Recognition
Sees: Reduction potential values for reducing strength ordering.
Shortcut: More negative standard reduction potential $\implies$ stronger reducing agent.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q71 (2026)
Consider the following electrochemical cell:
$$\text{Pt} \mid \text{O}_2(\text{g}) (1\text{ bar}) \mid \text{HCl}(\text{aq}) \mid\mid \text{M}^{2+}(\text{aq}, 1.0\text{ M}) \mid \text{M}(\text{s})$$
The pH above which, oxygen gas would start to evolve at anode is ____ (nearest integer).
Given: $E^0_{\text{M}^{2+}/\text{M}} = 0.994\text{ V}$, $E^0_{\text{O}_2/\text{H}_2\text{O}} = 1.23\text{ V}$ standard reduction potential and $\frac{RT}{F}(2.303) = 0.059\text{ V}$ at the given condition
### Related Formula
$$E_{\text{cell}} = E_{\text{Red}}(\text{cathode}) + E_{\text{Oxi}}(\text{anode}) > 0$$
$$E_{\text{Oxi}}(\text{anode}) = -E^0_{\text{O}_2/\text{H}_2\text{O}} - \frac{0.059}{2} \log \left([\text{H}^+]^2 P_{\text{O}_2}^{1/2}\right)$$
### Core Logic
Step 1: Set spontaneous cell reaction threshold ($E_{\text{cell}} = 0$):
$$E_{\text{Oxi}}(\text{anode}) = -E_{\text{Red}}(\text{cathode}) = -0.994\text{ V}$$
Step 2: Anodic reaction:
$$\text{H}_2\text{O} \rightarrow 2\text{H}^+ + \frac{1}{2}\text{O}_2 + 2e^-$$
$$E_{\text{Oxi}} = -1.23 - \frac{0.059}{2} \log \left([\text{H}^+]^2 \times 1^{1/2}\right) = -1.23 + 0.059 \times \text{pH}$$
Step 3: Equate oxidation potential:
$$-0.994 = -1.23 + 0.059 \times \text{pH}$$
$$0.059 \times \text{pH} = 0.236$$
$$\text{pH} = \frac{0.236}{0.059} = 4$$
### Pattern Recognition
Sees: Minimum pH for gas evolution at anode.
Shortcut: Apply $E_{\text{cell}} = 0$ threshold equation $-0.994 = -1.23 + 0.059 \text{pH} \implies \text{pH} = 4$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q55 (2026)
In the given electrochemical cell, $Ag(s)|AgCl(s)|Cl^{-}(aq) || FeCl_{2}(aq), FeCl_{3}(aq)|Pt(s)$ at 298 K, the cell potential ($E_{\text{cell}}$) will increase when :
(A) Concentration of $Fe^{2+}$ is increased.
(B) Concentration of $Fe^{3+}$ is decreased.
(C) Concentration of $Fe^{2+}$ is decreased.
(D) Concentration of $Fe^{3+}$ is increased.
(E) Concentration of $Cl^{-}$ is increased.
Choose the correct answer from the options given below :
- $\text{A and B only}$
- $\text{A and E only}$
- $\text{B only}$
- $\text{C, D and E only}$
### Related Formula
$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log Q$$
### Core Logic
First, write the complete balanced cell reaction.
At Anode (Oxidation): $Ag_{(s)} + Cl^-_{(aq)} \rightarrow AgCl_{(s)} + e^-$
At Cathode (Reduction): $Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$
Overall cell reaction:
$$Ag_{(s)} + Cl^-_{(aq)} + Fe^{3+}_{(aq)} \rightarrow AgCl_{(s)} + Fe^{2+}_{(aq)}$$
### Step 1: Applying the Nernst Equation
Applying the Nernst equation at $298 \text{ K}$ with $n = 1$:
$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{1} \log \frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$$
### Step 2: Analysis of Variables
To increase $E_{\text{cell}}$, the value of the logarithmic term $\log \frac{[Fe^{2+}]}{[Cl^-][Fe^{3+}]}$ must decrease.
This happens if the numerator decreases or the denominator increases.
- Decreasing $[Fe^{2+}]$ (Statement C)
- Increasing $[Fe^{3+}]$ (Statement D)
- Increasing $[Cl^-]$ (Statement E)
Thus, statements C, D, and E will increase the cell potential.
### Pattern Recognition
Nernst equation trick: $E_{\text{cell}}$ increases when product concentrations decrease or reactant concentrations increase (Le Chatelier's perspective of pushing the forward reaction).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
Q53 (2026)
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Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming $M^{+}$ ($M \rightarrow M^{+} + e^{-}$). The cation $M^{+}$ is present in two different concentrations $c_{1}$ and $c_{2}$ as shown above. Which of the following statement is correct for generating a positive cell potential?
- $\text{If } c_1 \text{ is present at anode, then } c_1 = c_2$
- $\text{If } c_1 \text{ is present at cathode, then } c_1 < c_2$
- $\text{If } c_1 \text{ is present at cathode, then } c_1 > c_2$
- $\text{If } c_1 \text{ is present at anode, then } c_1 > c_2$
### Related Formula
$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log Q$$
For a concentration cell, $E^{\circ}_{\text{cell}} = 0$, so:
$$E_{\text{cell}} = - \frac{0.0591}{n} \log \left( \frac{[\text{Anode}]}{[\text{Cathode}]} \right)$$
### Core Logic
For a concentration cell to have a positive cell potential ($E_{\text{cell}} > 0$), the ratio $\frac{[\text{Anode}]}{[\text{Cathode}]}$ must be less than $1$. This implies that $[\text{Anode}] < [\text{Cathode}]$.
Let's evaluate the given conditions:
**Case 1: If $c_1$ is at the anode.**
Then $c_2$ is at the cathode.
Cell reaction: $M(s) + M^{+}(c_2) \rightarrow M(s) + M^{+}(c_1)$
$$E_{\text{cell}} = -0.059 \log \frac{c_1}{c_2}$$
For $E_{\text{cell}} > 0$, we need $c_1 < c_2$.
Option 1 says $c_1 = c_2$ (Incorrect).
Option 4 says $c_1 > c_2$ (Incorrect).
### Step 1: Check Cathode Conditions
**Case 2: If $c_1$ is at the cathode.**
Then $c_2$ is at the anode.
Cell reaction: $M(s) + M^{+}(c_1) \rightarrow M(s) + M^{+}(c_2)$
$$E_{\text{cell}} = -0.059 \log \frac{c_2}{c_1}$$
For $E_{\text{cell}} > 0$, we need $\frac{c_2}{c_1} < 1 \implies c_2 < c_1 \implies c_1 > c_2$.
Option 2 says $c_1 < c_2$ (Incorrect).
Option 3 says $c_1 > c_2$ (Correct).
### Pattern Recognition
In any spontaneous concentration cell, ions flow from the higher concentration compartment to the lower concentration compartment. Thus, for a positive voltage, the cathode must always have the higher concentration.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Electrochemistry
