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JEE Main 2026 (Online) 21st January Evening Shift
Numerical
+4
-1
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MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.

$\mathrm{MX}(\mathrm{s}) \rightleftharpoons \mathrm{M}^{+}(\mathrm{aq})+\mathrm{X}^{-}(\mathrm{aq}) ; \quad \mathrm{K}_{\mathrm{sp}}=10^{-10}$

If the standard reduction potential for M+ (aq) + e → M(s) is
$\left(\mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\ominus}\right)=0.79 \mathrm{~V}$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $\mathrm{E}_{\mathrm{X}^{-} / \mathrm{MX}(\mathrm{s}) / \mathrm{M}}^{\ominus}$ is ______ mV. (nearest integer)

[Given: $ \dfrac{2.303 RT}{F} = 0.059\ \text{V} $]

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2
JEE Main 2026 (Online) 21st January Morning Shift
Numerical
+4
-1
Change Language

The pH and conductance of a weak acid $(\mathrm{HX})$ was found to be 5 and $4 \times 10^{-5} \mathrm{~S}$, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of $1 \mathrm{~cm}^2$ were at a distance of 15 cm apart. The value of the limiting molar conductivity is $\_\_\_\_$ $\mathrm{S} \mathrm{m}^2 \mathrm{~mol}^{-1}$. (nearest integer)

(Given : degree of dissociation of the weak acid $(\alpha) \ll 1$ )

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3
JEE Main 2025 (Online) 8th April Evening Shift
Numerical
+4
-1
Change Language

Consider the following half cell reaction

$$ \text{Cr}_2\text{O}_7^{2-} \, (\text{aq}) + 6\text{e}^- + 14\text{H}^+ \, (\text{aq}) \rightarrow 2\text{Cr}^{3+} \, (\text{aq}) + 7\text{H}_2\text{O} \, (\ell) $$

The reaction was conducted with the ratio of $$\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$$. The pH value at which the EMF of the half cell will become zero is __________.

(nearest integer value)

[Given: standard half cell reduction potential $$E^{\circ}_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\, \text{V}$$, $$\frac{2.303RT}{F} = 0.059\, \text{V}$$.]

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4
JEE Main 2025 (Online) 7th April Morning Shift
Numerical
+4
-1
Change Language

1 Faraday electricity was passed through $\mathrm{Cu}^{2+}(1.5 \mathrm{M}, 1 \mathrm{~L}) / \mathrm{Cu}$ and 0.1 Faraday was passed through $\mathrm{Ag}^{+}(0.2 \mathrm{M}, 1 \mathrm{~L}) / \mathrm{Ag}$ electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is __________ mV (nearest integer)

JEE Main 2025 (Online) 7th April Morning Shift Chemistry - Electrochemistry Question 26 English

$$\begin{aligned} \text { Given : } \mathrm{E}^{\circ} \mathrm{Cu}^{2+} / \mathrm{Cu} & =0.34 \mathrm{~V} \\\\ \mathrm{E}^{\circ} \mathrm{Ag}^{+} / \mathrm{Ag} & =0.8 \mathrm{~V} \\\\ \frac{2 \cdot 303 \mathrm{RT}}{\mathrm{~F}} & =0.06 \mathrm{~V} \end{aligned}$$

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