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

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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2
JEE Main 2025 (Online) 7th April Morning Shift
Numerical
+4
-1

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 5 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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3
JEE Main 2025 (Online) 2nd April Evening Shift
Numerical
+4
-1

$0.2 \%(\mathrm{w} / \mathrm{v})$ solution of NaOH is measured to have resistivity $870.0 \mathrm{~m} \Omega \mathrm{~m}$. The molar conductivity of the solution will be__________$\times 10^2 \mathrm{mS} \mathrm{dm}^2 \mathrm{~mol}^{-1}$. (Nearest integer)

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4
JEE Main 2025 (Online) 2nd April Morning Shift
Numerical
+4
-1

Consider the following electrochemical cell at standard condition.

$$\mathrm{Au}(\mathrm{~s})\left|\mathrm{QH}_2, \mathrm{Q}\right| \mathrm{NH}_4 \mathrm{X}(0.01 \mathrm{M})| | \mathrm{Ag}^{+}(1 \mathrm{M}) \mid \mathrm{Ag}(\mathrm{~s}) \mathrm{E}_{\text {cell }}=+0.4 \mathrm{~V}$$

The couple $\mathrm{QH}_2 / \mathrm{Q}$ represents quinhydrone electrode, the half cell reaction is given below:

JEE Main 2025 (Online) 2nd April Morning Shift Chemistry - Electrochemistry Question 8 English

$$\left[\text { Given : } \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=+0.8 \mathrm{~V} \text { and } \frac{2.303 \mathrm{RT}}{\mathrm{~F}}=0.06 \mathrm{~V}\right]$$

The $\mathrm{pK}_{\mathrm{b}}$ value of the ammonium halide salt $\left(\mathrm{NH}_4 \mathrm{X}\right)$ used here is __________ . (nearest integer)

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