JEE Main Chemistry: Electrochemistry High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27)

Correct Answer: d

Explanation: ​1.52 × 10$^{–4}$​ K<$_{a}$ = 1.69 × 10<$^{–5}$ Cα<$^{2}$ = 0.01α<$^{2}$ = 1.69 × 10<$^{–5}$ α = 0.04 Λ<$_{m}$ = α × Λ<$^{∞}$ = 0.04 × 380 × 10<$^{–4}$ Sm<$^{2}$ mol<$^{–1}$ = 15.2 S cm<$^{2}$ mol<$^{–1}$​ ​$N o w , \textrm{ }\textrm{ } \Lambda_{m} = \frac{\kappa \times 1000}{M}$​​ ​$o r \textrm{ }\textrm{ } \kappa = \frac{\Lambda_{m} \times M}{1000}$​​ ​$= \frac{15.2 \times 0.01}{1000}$​​ ​= 1.52 × 10<$^{–4}$ Scm<$^{–1}$​ ​1.52 × 10<$^{–2}$ Sm<$^{–1}$​

Correct Answer: d

Explanation: KCl $\begin{matrix}M o l e = \frac{Q}{n P} \\ F o r K C l \\ M o l e = \frac{1 F}{1 1 F} = 1 m o l e\end{matrix}$

Correct Answer: 2

Explanation: $\;\frac{W}{E}=\;\frac{\;\text{ charge }\;}{1\,\text{F}}$ $\;\frac{1.314}{\frac{197}{3}}=\;\frac{Q}{1\,\text{F}}$ $Q=2\times 10^{-2}\,\text{F}$

Correct Answer: (a)

Explanation: For the given reaction, $\begin{aligned} & \mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Pb}^{2+}(1 \mathrm{M}) \mid \mathrm{Pb}(s) \\ & E_{\mathrm{Zn}}^{\circ}=-0.763 \mathrm{~V} \text { and } E_{\mathrm{Pb}}^{\circ}=-0.126 \mathrm{~V} \\ & E_{\text {cell }}^{\circ}=? \end{aligned}$ Using formula, $\begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {cathode }}-E_{\text {anode }} \\ & =-0.126-(-0.763) \\ E_{\text {cell }}^{\circ} & =0.637 \mathrm{~V} \end{aligned}$

Correct Answer: 4

Explanation: $N i \&\text{nbsp}; \rightarrow \&\text{nbsp}; N i^{2 +}   + \&\text{nbsp}; 2 e^{-}$ $2 H^{+}   + \&\text{nbsp}; 2 e^{}   \rightarrow \&\text{nbsp}; H_{2}  $ $E_{c e l l}^{^{\text{o}}}   = \&\text{nbsp}; 0 . 236 \&\text{nbsp}; + \&\text{nbsp}; 0 \&\text{nbsp}; = \&\text{nbsp}; 0 . 236 \&\text{nbsp}; V$ $E_{c e l l}   = E_{c e l l}^{^{\text{o}}} - \frac{0 . 059}{2} log_{10} \frac{\left[N i^{2 +}\right]}{\left[H^{+}\right]^{2}}$ $0 = 0 . 236 + \frac{0 . 059}{2} log \frac{\left[H^{+}\right]^{2}}{1}  $ $log \left[H^{+}\right]^{2} = - \frac{0 . 236 \times 2}{0 . 059}  $ $2 log \left[H^{+}\right]   = - 8$ $- log \left[H^{+}\right] = \frac{8}{2} = 4$ $p H = 4$