275782
Find the emf of the following cell reaction, given $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\mathbf{0}}=-0.72 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{0}}^{\mathrm{o}}=-0.42 \mathrm{~V}$ $25^{\circ} \mathrm{C}$ is $\mathrm{Cr} \mid \mathrm{Cr}$
275788
In the Daniell cell, $\mathrm{Zn}\left \vert\mathrm{Zn}^{+} \ \vert \mathrm{Cu}^{2+}\right \vert \mathrm{Cu}$, when an external voltage is applied such that $E_{\text {external }}>$ $\mathbf{E}_{\text {cell }}$, current flows from
1 $\mathrm{Zn}$ to $\mathrm{Cu}$
2 $\mathrm{Cu}$ to $\mathrm{Zn}$
3 no current flows
4 data insufficient
Explanation:
On applying an external voltage greater than $1.1 \mathrm{~V}$ in a Daniell cell, the current flows in the reverse direction i.e., from $\mathrm{Zn}$ to $\mathrm{Cu}$ (cathode to anode) and electrons flow from $\mathrm{Cu}$ to $\mathrm{Zn}$. $\mathrm{Zn}$ is deposited at $\mathrm{Zn}$ electrode and $\mathrm{Cu}$ dissolves at $\mathrm{Cu}$ electrode. The reaction is $\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}$
According to Kohlrausch's law, Limiting molar conductivity of an electrolyte in the sum of the individual contributions of the cation and the anion of the electrolyte. Therefore, For option (a) $\begin{aligned} & \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \\ & =\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \end{aligned}$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS $\therefore \quad$ The LHS of option (a) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ and also RHS of that LHS $=$ RHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ For option (b) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Cl}^{-}\right)$ $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Br}^{-}\right)$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $=$ RHS Similarly for option (c) LHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{I}^{-}\right)$ and $\quad \mathrm{RHS}=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $\neq$ RHS And for option (d) also $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS Thus the correct answer is (c)
JEE Main 2020
ELECTROCHEMISTRY
275792
The correct statement about $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ among the following is (Given, atomic numbers of $\mathrm{Cr}=\mathbf{2 4}$ and $\mathrm{Mn}=25$ )
1 $\mathrm{Cr}^{2+}$ is a reducing agent
2 $\mathrm{Mn}^{3+}$ is a reducing agent
3 Both $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ exhibit $\mathrm{d}^{4}$ outer electronic configuration
4 When $\mathrm{Cr}^{2+}$ is used as reducing agent, it attains $\mathrm{d}^{5}$ electronic configuration
Explanation:
$\mathrm{Cr}^{2+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{4}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ}=-0.41 \mathrm{~V}$ $\mathrm{Mn}^{3+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{3}, \mathrm{E}_{\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}}^{\circ}=1.51 \mathrm{~V}$ So, reducing power $\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}$
AP-EAMCET (Engg.) 17.09.2020 Shift-I
ELECTROCHEMISTRY
275798
If electrolysis of aqueous $\mathrm{CuSO}_{4}$ solution is carried out using $\mathrm{Cu}$-electrodes, the reaction taking place at the anode is
$\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}$At anode (oxidation) $\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \quad \text { At cathode }$ Copper dissolves into the solution from the anode and it deposited at the cathode.
275782
Find the emf of the following cell reaction, given $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\mathbf{0}}=-0.72 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{0}}^{\mathrm{o}}=-0.42 \mathrm{~V}$ $25^{\circ} \mathrm{C}$ is $\mathrm{Cr} \mid \mathrm{Cr}$
275788
In the Daniell cell, $\mathrm{Zn}\left \vert\mathrm{Zn}^{+} \ \vert \mathrm{Cu}^{2+}\right \vert \mathrm{Cu}$, when an external voltage is applied such that $E_{\text {external }}>$ $\mathbf{E}_{\text {cell }}$, current flows from
1 $\mathrm{Zn}$ to $\mathrm{Cu}$
2 $\mathrm{Cu}$ to $\mathrm{Zn}$
3 no current flows
4 data insufficient
Explanation:
On applying an external voltage greater than $1.1 \mathrm{~V}$ in a Daniell cell, the current flows in the reverse direction i.e., from $\mathrm{Zn}$ to $\mathrm{Cu}$ (cathode to anode) and electrons flow from $\mathrm{Cu}$ to $\mathrm{Zn}$. $\mathrm{Zn}$ is deposited at $\mathrm{Zn}$ electrode and $\mathrm{Cu}$ dissolves at $\mathrm{Cu}$ electrode. The reaction is $\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}$
According to Kohlrausch's law, Limiting molar conductivity of an electrolyte in the sum of the individual contributions of the cation and the anion of the electrolyte. Therefore, For option (a) $\begin{aligned} & \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \\ & =\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \end{aligned}$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS $\therefore \quad$ The LHS of option (a) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ and also RHS of that LHS $=$ RHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ For option (b) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Cl}^{-}\right)$ $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Br}^{-}\right)$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $=$ RHS Similarly for option (c) LHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{I}^{-}\right)$ and $\quad \mathrm{RHS}=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $\neq$ RHS And for option (d) also $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS Thus the correct answer is (c)
JEE Main 2020
ELECTROCHEMISTRY
275792
The correct statement about $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ among the following is (Given, atomic numbers of $\mathrm{Cr}=\mathbf{2 4}$ and $\mathrm{Mn}=25$ )
1 $\mathrm{Cr}^{2+}$ is a reducing agent
2 $\mathrm{Mn}^{3+}$ is a reducing agent
3 Both $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ exhibit $\mathrm{d}^{4}$ outer electronic configuration
4 When $\mathrm{Cr}^{2+}$ is used as reducing agent, it attains $\mathrm{d}^{5}$ electronic configuration
Explanation:
$\mathrm{Cr}^{2+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{4}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ}=-0.41 \mathrm{~V}$ $\mathrm{Mn}^{3+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{3}, \mathrm{E}_{\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}}^{\circ}=1.51 \mathrm{~V}$ So, reducing power $\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}$
AP-EAMCET (Engg.) 17.09.2020 Shift-I
ELECTROCHEMISTRY
275798
If electrolysis of aqueous $\mathrm{CuSO}_{4}$ solution is carried out using $\mathrm{Cu}$-electrodes, the reaction taking place at the anode is
$\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}$At anode (oxidation) $\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \quad \text { At cathode }$ Copper dissolves into the solution from the anode and it deposited at the cathode.
275782
Find the emf of the following cell reaction, given $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\mathbf{0}}=-0.72 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{0}}^{\mathrm{o}}=-0.42 \mathrm{~V}$ $25^{\circ} \mathrm{C}$ is $\mathrm{Cr} \mid \mathrm{Cr}$
275788
In the Daniell cell, $\mathrm{Zn}\left \vert\mathrm{Zn}^{+} \ \vert \mathrm{Cu}^{2+}\right \vert \mathrm{Cu}$, when an external voltage is applied such that $E_{\text {external }}>$ $\mathbf{E}_{\text {cell }}$, current flows from
1 $\mathrm{Zn}$ to $\mathrm{Cu}$
2 $\mathrm{Cu}$ to $\mathrm{Zn}$
3 no current flows
4 data insufficient
Explanation:
On applying an external voltage greater than $1.1 \mathrm{~V}$ in a Daniell cell, the current flows in the reverse direction i.e., from $\mathrm{Zn}$ to $\mathrm{Cu}$ (cathode to anode) and electrons flow from $\mathrm{Cu}$ to $\mathrm{Zn}$. $\mathrm{Zn}$ is deposited at $\mathrm{Zn}$ electrode and $\mathrm{Cu}$ dissolves at $\mathrm{Cu}$ electrode. The reaction is $\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}$
According to Kohlrausch's law, Limiting molar conductivity of an electrolyte in the sum of the individual contributions of the cation and the anion of the electrolyte. Therefore, For option (a) $\begin{aligned} & \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \\ & =\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \end{aligned}$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS $\therefore \quad$ The LHS of option (a) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ and also RHS of that LHS $=$ RHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ For option (b) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Cl}^{-}\right)$ $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Br}^{-}\right)$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $=$ RHS Similarly for option (c) LHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{I}^{-}\right)$ and $\quad \mathrm{RHS}=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $\neq$ RHS And for option (d) also $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS Thus the correct answer is (c)
JEE Main 2020
ELECTROCHEMISTRY
275792
The correct statement about $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ among the following is (Given, atomic numbers of $\mathrm{Cr}=\mathbf{2 4}$ and $\mathrm{Mn}=25$ )
1 $\mathrm{Cr}^{2+}$ is a reducing agent
2 $\mathrm{Mn}^{3+}$ is a reducing agent
3 Both $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ exhibit $\mathrm{d}^{4}$ outer electronic configuration
4 When $\mathrm{Cr}^{2+}$ is used as reducing agent, it attains $\mathrm{d}^{5}$ electronic configuration
Explanation:
$\mathrm{Cr}^{2+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{4}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ}=-0.41 \mathrm{~V}$ $\mathrm{Mn}^{3+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{3}, \mathrm{E}_{\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}}^{\circ}=1.51 \mathrm{~V}$ So, reducing power $\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}$
AP-EAMCET (Engg.) 17.09.2020 Shift-I
ELECTROCHEMISTRY
275798
If electrolysis of aqueous $\mathrm{CuSO}_{4}$ solution is carried out using $\mathrm{Cu}$-electrodes, the reaction taking place at the anode is
$\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}$At anode (oxidation) $\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \quad \text { At cathode }$ Copper dissolves into the solution from the anode and it deposited at the cathode.
275782
Find the emf of the following cell reaction, given $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\mathbf{0}}=-0.72 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{0}}^{\mathrm{o}}=-0.42 \mathrm{~V}$ $25^{\circ} \mathrm{C}$ is $\mathrm{Cr} \mid \mathrm{Cr}$
275788
In the Daniell cell, $\mathrm{Zn}\left \vert\mathrm{Zn}^{+} \ \vert \mathrm{Cu}^{2+}\right \vert \mathrm{Cu}$, when an external voltage is applied such that $E_{\text {external }}>$ $\mathbf{E}_{\text {cell }}$, current flows from
1 $\mathrm{Zn}$ to $\mathrm{Cu}$
2 $\mathrm{Cu}$ to $\mathrm{Zn}$
3 no current flows
4 data insufficient
Explanation:
On applying an external voltage greater than $1.1 \mathrm{~V}$ in a Daniell cell, the current flows in the reverse direction i.e., from $\mathrm{Zn}$ to $\mathrm{Cu}$ (cathode to anode) and electrons flow from $\mathrm{Cu}$ to $\mathrm{Zn}$. $\mathrm{Zn}$ is deposited at $\mathrm{Zn}$ electrode and $\mathrm{Cu}$ dissolves at $\mathrm{Cu}$ electrode. The reaction is $\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}$
According to Kohlrausch's law, Limiting molar conductivity of an electrolyte in the sum of the individual contributions of the cation and the anion of the electrolyte. Therefore, For option (a) $\begin{aligned} & \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \\ & =\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \end{aligned}$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS $\therefore \quad$ The LHS of option (a) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ and also RHS of that LHS $=$ RHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ For option (b) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Cl}^{-}\right)$ $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Br}^{-}\right)$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $=$ RHS Similarly for option (c) LHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{I}^{-}\right)$ and $\quad \mathrm{RHS}=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $\neq$ RHS And for option (d) also $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS Thus the correct answer is (c)
JEE Main 2020
ELECTROCHEMISTRY
275792
The correct statement about $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ among the following is (Given, atomic numbers of $\mathrm{Cr}=\mathbf{2 4}$ and $\mathrm{Mn}=25$ )
1 $\mathrm{Cr}^{2+}$ is a reducing agent
2 $\mathrm{Mn}^{3+}$ is a reducing agent
3 Both $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ exhibit $\mathrm{d}^{4}$ outer electronic configuration
4 When $\mathrm{Cr}^{2+}$ is used as reducing agent, it attains $\mathrm{d}^{5}$ electronic configuration
Explanation:
$\mathrm{Cr}^{2+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{4}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ}=-0.41 \mathrm{~V}$ $\mathrm{Mn}^{3+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{3}, \mathrm{E}_{\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}}^{\circ}=1.51 \mathrm{~V}$ So, reducing power $\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}$
AP-EAMCET (Engg.) 17.09.2020 Shift-I
ELECTROCHEMISTRY
275798
If electrolysis of aqueous $\mathrm{CuSO}_{4}$ solution is carried out using $\mathrm{Cu}$-electrodes, the reaction taking place at the anode is
$\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}$At anode (oxidation) $\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \quad \text { At cathode }$ Copper dissolves into the solution from the anode and it deposited at the cathode.
275782
Find the emf of the following cell reaction, given $\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\mathbf{0}}=-0.72 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{0}}^{\mathrm{o}}=-0.42 \mathrm{~V}$ $25^{\circ} \mathrm{C}$ is $\mathrm{Cr} \mid \mathrm{Cr}$
275788
In the Daniell cell, $\mathrm{Zn}\left \vert\mathrm{Zn}^{+} \ \vert \mathrm{Cu}^{2+}\right \vert \mathrm{Cu}$, when an external voltage is applied such that $E_{\text {external }}>$ $\mathbf{E}_{\text {cell }}$, current flows from
1 $\mathrm{Zn}$ to $\mathrm{Cu}$
2 $\mathrm{Cu}$ to $\mathrm{Zn}$
3 no current flows
4 data insufficient
Explanation:
On applying an external voltage greater than $1.1 \mathrm{~V}$ in a Daniell cell, the current flows in the reverse direction i.e., from $\mathrm{Zn}$ to $\mathrm{Cu}$ (cathode to anode) and electrons flow from $\mathrm{Cu}$ to $\mathrm{Zn}$. $\mathrm{Zn}$ is deposited at $\mathrm{Zn}$ electrode and $\mathrm{Cu}$ dissolves at $\mathrm{Cu}$ electrode. The reaction is $\mathrm{Zn}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Zn}+\mathrm{Cu}^{2+}$
According to Kohlrausch's law, Limiting molar conductivity of an electrolyte in the sum of the individual contributions of the cation and the anion of the electrolyte. Therefore, For option (a) $\begin{aligned} & \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \\ & =\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right) \end{aligned}$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS $\therefore \quad$ The LHS of option (a) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ and also RHS of that LHS $=$ RHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)$ For option (b) $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Cl}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Cl}^{-}\right)$ $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Br}^{-}\right)$ or $\quad \lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda^{\circ} \mathrm{m}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $=$ RHS Similarly for option (c) LHS $=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Br}^{-}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{I}^{-}\right)$ and $\quad \mathrm{RHS}=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{K}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{Na}^{+}\right)$ $\therefore \quad$ LHS $\neq$ RHS And for option (d) also $\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)-\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)=\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{H}^{+}\right)+\lambda_{\mathrm{m}}^{\circ}\left(\mathrm{OH}^{-}\right)$ $\therefore \quad$ LHS $=$ RHS Thus the correct answer is (c)
JEE Main 2020
ELECTROCHEMISTRY
275792
The correct statement about $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ among the following is (Given, atomic numbers of $\mathrm{Cr}=\mathbf{2 4}$ and $\mathrm{Mn}=25$ )
1 $\mathrm{Cr}^{2+}$ is a reducing agent
2 $\mathrm{Mn}^{3+}$ is a reducing agent
3 Both $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ exhibit $\mathrm{d}^{4}$ outer electronic configuration
4 When $\mathrm{Cr}^{2+}$ is used as reducing agent, it attains $\mathrm{d}^{5}$ electronic configuration
Explanation:
$\mathrm{Cr}^{2+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{4}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\circ}=-0.41 \mathrm{~V}$ $\mathrm{Mn}^{3+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{3}, \mathrm{E}_{\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}}^{\circ}=1.51 \mathrm{~V}$ So, reducing power $\mathrm{Cr}^{2+}>\mathrm{Mn}^{2+}$
AP-EAMCET (Engg.) 17.09.2020 Shift-I
ELECTROCHEMISTRY
275798
If electrolysis of aqueous $\mathrm{CuSO}_{4}$ solution is carried out using $\mathrm{Cu}$-electrodes, the reaction taking place at the anode is
$\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}$At anode (oxidation) $\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu} \quad \text { At cathode }$ Copper dissolves into the solution from the anode and it deposited at the cathode.
