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Electrochemistry question

2025 · 7 Apr · Shift 1 · Q23
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Electrochemistry question

2025 · 7 Apr · Shift 1 · Q23

JEE MainChemistryElectrochemistryNumerical+4 / −1
1 Faraday electricity was passed through Cu2+(1.5M,1 L)/Cu\mathrm{Cu}^{2+}(1.5 \mathrm{M}, 1 \mathrm{~L}) / \mathrm{Cu}Cu2+(1.5M,1 L)/Cu and 0.1 Faraday was passed through Ag+(0.2M,1 L)/Ag\mathrm{Ag}^{+}(0.2 \mathrm{M}, 1 \mathrm{~L}) / \mathrm{Ag}Ag+(0.2M,1 L)/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 ‾\underline{\hspace{2cm}}​ mV (nearest integer) JEE Main 2025 (Online) 7th April Morning Shift Chemistry - Electrochemistry Question 5 English Given : E∘Cu2+/Cu=0.34 VE∘Ag+/Ag=0.8 V2⋅303RT F=0.06 V\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} Given : E∘Cu2+/CuE∘Ag+/Ag F2⋅303RT​​=0.34 V=0.8 V=0.06 V​
Numerical answer
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