JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
\(1\) 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-

Given: \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}=0.34 \mathrm{~V}\)
\(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=0.8 \mathrm{~V}\)
\(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}\)
- A 100
- B 200
- C 300
- D 400
Answer & Solution
Correct Answer
(D) 400
Step-by-step Solution
Detailed explanation
\(\mathrm{Cu}^{+2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}\)…
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