1 Faraday electricity was passed through Cu^ 2+ (1.5 M, 1 ~L) / Cu and 0.1 Faraday was passed through Ag^ + (0.2 M, 1 ~L) / Ag electrolytic cells. After this the two cells were connected as shown below to make an

Correct answer: 400 * Cu^ +2 +2 e^ - Cu aligned & (1 faraday = charge on 1 mole electron ) \& t=0 1.5 1 ~mole \& t=t 1 - 0.5 mole aligned [Cu^ +2 ]=1 M after electrolysis * Ag +e^ - Ag array lll t=0 & 0.2 & 0.1 mole t=t & 0.1 & - - array [Ag^ + ]=0.1 M after electrolysis Cell Cu_ (s) +2 Ag_ (aq) ^ + arrow Cu_ (aq) ^ +2 +2 Ag_ (s) reaction aligned & E=E^ - 0.06 n [Cu^ +2 ] [Ag^ + ]^2 & E=(0.8-0.34)-(0.06)/(2) (1)/((0.1)^2)=0.4 ~V aligned Correct answer =400 mV

JEE Main 20257 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Electrochemistry

JEE Main 2025 · 7 April · Morning Shift · Chemistry · Electrochemistry

1 Faraday electricity was passed through Cu^ 2+ (1.5 M, 1 ~L) / Cu...

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-
1 Faraday electricity was passed through Cu^ 2+ (1.5 M, 1 ~L) / Cu and 0.1 Faraday was passed through Ag^ + (0.2 M, 1

1 Faraday electricity was passed through Cu^ 2+ (1.5 M, 1 ~L) / Cu and 0.1 Faraday was passed through Ag^ + (0.2 M, 1

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}

Correct answer

400

Step-by-step solution

\mathrm{Cu}^{+2}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}

\begin{aligned}& (1 \text{ faraday }=\text{ charge on } 1 \text{ mole electron }) \& \mathrm{t}=0 \quad 1.5 \quad 1 \mathrm{~mole} \& \mathrm{t}=\mathrm{t} \quad 1 \quad-\quad 0.5 \text{ mole }\end{aligned}

\left[\mathrm{Cu}^{+2}\right]=1 \mathrm{M}

after electrolysis<br />*

\mathrm{Ag} \oplus+\mathrm{e}^{-} \quad \longrightarrow \mathrm{Ag}

<br />

\begin{array}{lll}\mathrm{t}=0 & 0.2 & 0.1 \text{ mole } \\ \mathrm{t}=\mathrm{t} & 0.1 & - \qquad -\end{array}

<br />\left[\mathrm{Ag}^{+}\right]=0.1 \mathrm{M}

after electrolysis
Cell

\quad \mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\mathrm{aq})}^{+} \rightarrow \mathrm{Cu}_{(\mathrm{aq})}^{+2}+2 \mathrm{Ag}_{(\mathrm{s})}

reaction

\begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.06}{\mathrm{n}} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \\ & \mathrm{E}=(0.8-0.34)-\frac{0.06}{2} \log \frac{1}{(0.1)^2}=0.4 \mathrm{~V}\end{aligned}

Correct answer

=400 \mathrm{mV}

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