275757
Half-life of a reaction is found to be inversely proportional to the fifth power is initial concentration, the order of reaction is
1 5
2 6
3 3
4 4
Explanation:
$\mathrm{t}_{1 / 2} \propto \frac{1}{\mathrm{a}^{\mathrm{n}-1}}, \mathrm{n}=6$ $\begin{aligned} & t_{1 / 2}=\frac{1}{a^{6-1}} \\ & t_{1 / 2}=\frac{1}{a^{5}} \end{aligned}$ Hence, the order of reaction is 6 .
Karnataka CET-17.06.2022
ELECTROCHEMISTRY
275758
The correct order of reduction potentials of the following pairs is
1 A $>$ C $>$ B $>$ D $>$ E
2 A $>$ B $>$ C $>$ D $>$ E
3 A $>$ C $>$ B $>$ E $>$ D
4 A $>$ B $>$ C $>$ E $>$ D
5 $\mathrm{Ag}^{+} / \mathrm{Ag}$ Choose the correct answer from the options given below:
Explanation:
(A) $\mathrm{E}_{\mathrm{Cl}_{2} / \mathrm{Cl}^{-}}^{\mathrm{o}}=1.36 \mathrm{~V}$ $\begin{aligned} & \text { (B) } \mathrm{E}_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{\mathrm{o}}=0.54 \mathrm{~V} \\ & \text { (C) } \mathrm{Eg}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}=0.80 \mathrm{~V} \\ & \text { (D) } \mathrm{E}_{\mathrm{Na}^{+} / \mathrm{Na}}^{\mathrm{o}}=-2.71 \mathrm{~V} \\ & \text { (E) } \mathrm{E}_{\mathrm{Li}^{+} / \mathrm{Li}}^{\mathrm{o}}=-3.05 \mathrm{~V} \end{aligned}$ The correct order of reduction potential of the following pairs is $\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}>\mathrm{E}$
JEE Main-25.06.2022
ELECTROCHEMISTRY
275759
96.5 amperes current is passed through the molten $\mathrm{AlCl}_{3}$ for 100 seconds. The mass of aluminum deposited a the cathode is (Atomic weight of $\mathrm{Al}$ - $27 \mathrm{u}$ )
1 $0.90 \mathrm{~g}$
2 $0.45 \mathrm{~g}$
3 $1.35 \mathrm{~g}$
4 $1.8 \mathrm{~g}$
Explanation:
Given, current (i) $=96.5$ amperes Time $(t)=100$ second $\begin{aligned} & \mathrm{m}=2 \\ & \mathrm{E}=27 \end{aligned}$ Faraday second law, $\begin{aligned} \mathrm{m} & =\frac{\mathrm{E}}{\mathrm{F}} \times \text { it } \\ \mathrm{m} & =\frac{27}{3 \times 96500} \times 96.5 \times 100 \\ \mathrm{~m} & =0.9 \mathrm{~g} \end{aligned}$
Shift-I
ELECTROCHEMISTRY
275760
The $\mathrm{E}^{0}$ of $\mathrm{Ce}^{4+} / \mathrm{Ce}^{+3}=1.6 \mathrm{~V}$ $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.76 \mathrm{~V}$ the $\mathrm{E}^{\mathbf{0}}$ of $\mathrm{Fe}^{3+}$ oxidising $\mathrm{Ce}^{3+}$ is
1 $+0.84 \mathrm{~V}$
2 $-0.84 \mathrm{~V}$
3 $-2.32 \mathrm{~V}$
4 $+1.5 \mathrm{~V}$
Explanation:
Given that, $\begin{aligned} & \mathrm{E}^{\mathrm{o}} \text { of } \mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=1.6 \mathrm{~V} \\ & \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.76 \mathrm{~V} \end{aligned}$ To find $\mathrm{E}^{\mathrm{o}}$ of $\mathrm{Fe}^{3+}$ oxidising $\mathrm{Ce}^{3+}$ is - $\mathrm{E}^{\mathrm{o}}=$ Reduction potential of cathode - Reduction Potential of anode $\begin{aligned} & \mathrm{E}^{\mathrm{o}}=\left(\mathrm{Fe}^{3+}\right)_{\operatorname{Red}}-\left(\mathrm{Ce}^{3+}\right)_{\operatorname{Red}} \\ &= 0.76-1.6 \\ & \mathrm{E}^{\mathrm{o}}=-0.84 \mathrm{~V} \end{aligned}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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ELECTROCHEMISTRY
275757
Half-life of a reaction is found to be inversely proportional to the fifth power is initial concentration, the order of reaction is
1 5
2 6
3 3
4 4
Explanation:
$\mathrm{t}_{1 / 2} \propto \frac{1}{\mathrm{a}^{\mathrm{n}-1}}, \mathrm{n}=6$ $\begin{aligned} & t_{1 / 2}=\frac{1}{a^{6-1}} \\ & t_{1 / 2}=\frac{1}{a^{5}} \end{aligned}$ Hence, the order of reaction is 6 .
Karnataka CET-17.06.2022
ELECTROCHEMISTRY
275758
The correct order of reduction potentials of the following pairs is
1 A $>$ C $>$ B $>$ D $>$ E
2 A $>$ B $>$ C $>$ D $>$ E
3 A $>$ C $>$ B $>$ E $>$ D
4 A $>$ B $>$ C $>$ E $>$ D
5 $\mathrm{Ag}^{+} / \mathrm{Ag}$ Choose the correct answer from the options given below:
Explanation:
(A) $\mathrm{E}_{\mathrm{Cl}_{2} / \mathrm{Cl}^{-}}^{\mathrm{o}}=1.36 \mathrm{~V}$ $\begin{aligned} & \text { (B) } \mathrm{E}_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{\mathrm{o}}=0.54 \mathrm{~V} \\ & \text { (C) } \mathrm{Eg}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}=0.80 \mathrm{~V} \\ & \text { (D) } \mathrm{E}_{\mathrm{Na}^{+} / \mathrm{Na}}^{\mathrm{o}}=-2.71 \mathrm{~V} \\ & \text { (E) } \mathrm{E}_{\mathrm{Li}^{+} / \mathrm{Li}}^{\mathrm{o}}=-3.05 \mathrm{~V} \end{aligned}$ The correct order of reduction potential of the following pairs is $\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}>\mathrm{E}$
JEE Main-25.06.2022
ELECTROCHEMISTRY
275759
96.5 amperes current is passed through the molten $\mathrm{AlCl}_{3}$ for 100 seconds. The mass of aluminum deposited a the cathode is (Atomic weight of $\mathrm{Al}$ - $27 \mathrm{u}$ )
1 $0.90 \mathrm{~g}$
2 $0.45 \mathrm{~g}$
3 $1.35 \mathrm{~g}$
4 $1.8 \mathrm{~g}$
Explanation:
Given, current (i) $=96.5$ amperes Time $(t)=100$ second $\begin{aligned} & \mathrm{m}=2 \\ & \mathrm{E}=27 \end{aligned}$ Faraday second law, $\begin{aligned} \mathrm{m} & =\frac{\mathrm{E}}{\mathrm{F}} \times \text { it } \\ \mathrm{m} & =\frac{27}{3 \times 96500} \times 96.5 \times 100 \\ \mathrm{~m} & =0.9 \mathrm{~g} \end{aligned}$
Shift-I
ELECTROCHEMISTRY
275760
The $\mathrm{E}^{0}$ of $\mathrm{Ce}^{4+} / \mathrm{Ce}^{+3}=1.6 \mathrm{~V}$ $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.76 \mathrm{~V}$ the $\mathrm{E}^{\mathbf{0}}$ of $\mathrm{Fe}^{3+}$ oxidising $\mathrm{Ce}^{3+}$ is
1 $+0.84 \mathrm{~V}$
2 $-0.84 \mathrm{~V}$
3 $-2.32 \mathrm{~V}$
4 $+1.5 \mathrm{~V}$
Explanation:
Given that, $\begin{aligned} & \mathrm{E}^{\mathrm{o}} \text { of } \mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=1.6 \mathrm{~V} \\ & \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.76 \mathrm{~V} \end{aligned}$ To find $\mathrm{E}^{\mathrm{o}}$ of $\mathrm{Fe}^{3+}$ oxidising $\mathrm{Ce}^{3+}$ is - $\mathrm{E}^{\mathrm{o}}=$ Reduction potential of cathode - Reduction Potential of anode $\begin{aligned} & \mathrm{E}^{\mathrm{o}}=\left(\mathrm{Fe}^{3+}\right)_{\operatorname{Red}}-\left(\mathrm{Ce}^{3+}\right)_{\operatorname{Red}} \\ &= 0.76-1.6 \\ & \mathrm{E}^{\mathrm{o}}=-0.84 \mathrm{~V} \end{aligned}$
275757
Half-life of a reaction is found to be inversely proportional to the fifth power is initial concentration, the order of reaction is
1 5
2 6
3 3
4 4
Explanation:
$\mathrm{t}_{1 / 2} \propto \frac{1}{\mathrm{a}^{\mathrm{n}-1}}, \mathrm{n}=6$ $\begin{aligned} & t_{1 / 2}=\frac{1}{a^{6-1}} \\ & t_{1 / 2}=\frac{1}{a^{5}} \end{aligned}$ Hence, the order of reaction is 6 .
Karnataka CET-17.06.2022
ELECTROCHEMISTRY
275758
The correct order of reduction potentials of the following pairs is
1 A $>$ C $>$ B $>$ D $>$ E
2 A $>$ B $>$ C $>$ D $>$ E
3 A $>$ C $>$ B $>$ E $>$ D
4 A $>$ B $>$ C $>$ E $>$ D
5 $\mathrm{Ag}^{+} / \mathrm{Ag}$ Choose the correct answer from the options given below:
Explanation:
(A) $\mathrm{E}_{\mathrm{Cl}_{2} / \mathrm{Cl}^{-}}^{\mathrm{o}}=1.36 \mathrm{~V}$ $\begin{aligned} & \text { (B) } \mathrm{E}_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{\mathrm{o}}=0.54 \mathrm{~V} \\ & \text { (C) } \mathrm{Eg}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}=0.80 \mathrm{~V} \\ & \text { (D) } \mathrm{E}_{\mathrm{Na}^{+} / \mathrm{Na}}^{\mathrm{o}}=-2.71 \mathrm{~V} \\ & \text { (E) } \mathrm{E}_{\mathrm{Li}^{+} / \mathrm{Li}}^{\mathrm{o}}=-3.05 \mathrm{~V} \end{aligned}$ The correct order of reduction potential of the following pairs is $\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}>\mathrm{E}$
JEE Main-25.06.2022
ELECTROCHEMISTRY
275759
96.5 amperes current is passed through the molten $\mathrm{AlCl}_{3}$ for 100 seconds. The mass of aluminum deposited a the cathode is (Atomic weight of $\mathrm{Al}$ - $27 \mathrm{u}$ )
1 $0.90 \mathrm{~g}$
2 $0.45 \mathrm{~g}$
3 $1.35 \mathrm{~g}$
4 $1.8 \mathrm{~g}$
Explanation:
Given, current (i) $=96.5$ amperes Time $(t)=100$ second $\begin{aligned} & \mathrm{m}=2 \\ & \mathrm{E}=27 \end{aligned}$ Faraday second law, $\begin{aligned} \mathrm{m} & =\frac{\mathrm{E}}{\mathrm{F}} \times \text { it } \\ \mathrm{m} & =\frac{27}{3 \times 96500} \times 96.5 \times 100 \\ \mathrm{~m} & =0.9 \mathrm{~g} \end{aligned}$
Shift-I
ELECTROCHEMISTRY
275760
The $\mathrm{E}^{0}$ of $\mathrm{Ce}^{4+} / \mathrm{Ce}^{+3}=1.6 \mathrm{~V}$ $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.76 \mathrm{~V}$ the $\mathrm{E}^{\mathbf{0}}$ of $\mathrm{Fe}^{3+}$ oxidising $\mathrm{Ce}^{3+}$ is
1 $+0.84 \mathrm{~V}$
2 $-0.84 \mathrm{~V}$
3 $-2.32 \mathrm{~V}$
4 $+1.5 \mathrm{~V}$
Explanation:
Given that, $\begin{aligned} & \mathrm{E}^{\mathrm{o}} \text { of } \mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=1.6 \mathrm{~V} \\ & \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.76 \mathrm{~V} \end{aligned}$ To find $\mathrm{E}^{\mathrm{o}}$ of $\mathrm{Fe}^{3+}$ oxidising $\mathrm{Ce}^{3+}$ is - $\mathrm{E}^{\mathrm{o}}=$ Reduction potential of cathode - Reduction Potential of anode $\begin{aligned} & \mathrm{E}^{\mathrm{o}}=\left(\mathrm{Fe}^{3+}\right)_{\operatorname{Red}}-\left(\mathrm{Ce}^{3+}\right)_{\operatorname{Red}} \\ &= 0.76-1.6 \\ & \mathrm{E}^{\mathrm{o}}=-0.84 \mathrm{~V} \end{aligned}$
275757
Half-life of a reaction is found to be inversely proportional to the fifth power is initial concentration, the order of reaction is
1 5
2 6
3 3
4 4
Explanation:
$\mathrm{t}_{1 / 2} \propto \frac{1}{\mathrm{a}^{\mathrm{n}-1}}, \mathrm{n}=6$ $\begin{aligned} & t_{1 / 2}=\frac{1}{a^{6-1}} \\ & t_{1 / 2}=\frac{1}{a^{5}} \end{aligned}$ Hence, the order of reaction is 6 .
Karnataka CET-17.06.2022
ELECTROCHEMISTRY
275758
The correct order of reduction potentials of the following pairs is
1 A $>$ C $>$ B $>$ D $>$ E
2 A $>$ B $>$ C $>$ D $>$ E
3 A $>$ C $>$ B $>$ E $>$ D
4 A $>$ B $>$ C $>$ E $>$ D
5 $\mathrm{Ag}^{+} / \mathrm{Ag}$ Choose the correct answer from the options given below:
Explanation:
(A) $\mathrm{E}_{\mathrm{Cl}_{2} / \mathrm{Cl}^{-}}^{\mathrm{o}}=1.36 \mathrm{~V}$ $\begin{aligned} & \text { (B) } \mathrm{E}_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{\mathrm{o}}=0.54 \mathrm{~V} \\ & \text { (C) } \mathrm{Eg}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}}=0.80 \mathrm{~V} \\ & \text { (D) } \mathrm{E}_{\mathrm{Na}^{+} / \mathrm{Na}}^{\mathrm{o}}=-2.71 \mathrm{~V} \\ & \text { (E) } \mathrm{E}_{\mathrm{Li}^{+} / \mathrm{Li}}^{\mathrm{o}}=-3.05 \mathrm{~V} \end{aligned}$ The correct order of reduction potential of the following pairs is $\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}>\mathrm{E}$
JEE Main-25.06.2022
ELECTROCHEMISTRY
275759
96.5 amperes current is passed through the molten $\mathrm{AlCl}_{3}$ for 100 seconds. The mass of aluminum deposited a the cathode is (Atomic weight of $\mathrm{Al}$ - $27 \mathrm{u}$ )
1 $0.90 \mathrm{~g}$
2 $0.45 \mathrm{~g}$
3 $1.35 \mathrm{~g}$
4 $1.8 \mathrm{~g}$
Explanation:
Given, current (i) $=96.5$ amperes Time $(t)=100$ second $\begin{aligned} & \mathrm{m}=2 \\ & \mathrm{E}=27 \end{aligned}$ Faraday second law, $\begin{aligned} \mathrm{m} & =\frac{\mathrm{E}}{\mathrm{F}} \times \text { it } \\ \mathrm{m} & =\frac{27}{3 \times 96500} \times 96.5 \times 100 \\ \mathrm{~m} & =0.9 \mathrm{~g} \end{aligned}$
Shift-I
ELECTROCHEMISTRY
275760
The $\mathrm{E}^{0}$ of $\mathrm{Ce}^{4+} / \mathrm{Ce}^{+3}=1.6 \mathrm{~V}$ $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.76 \mathrm{~V}$ the $\mathrm{E}^{\mathbf{0}}$ of $\mathrm{Fe}^{3+}$ oxidising $\mathrm{Ce}^{3+}$ is
1 $+0.84 \mathrm{~V}$
2 $-0.84 \mathrm{~V}$
3 $-2.32 \mathrm{~V}$
4 $+1.5 \mathrm{~V}$
Explanation:
Given that, $\begin{aligned} & \mathrm{E}^{\mathrm{o}} \text { of } \mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=1.6 \mathrm{~V} \\ & \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.76 \mathrm{~V} \end{aligned}$ To find $\mathrm{E}^{\mathrm{o}}$ of $\mathrm{Fe}^{3+}$ oxidising $\mathrm{Ce}^{3+}$ is - $\mathrm{E}^{\mathrm{o}}=$ Reduction potential of cathode - Reduction Potential of anode $\begin{aligned} & \mathrm{E}^{\mathrm{o}}=\left(\mathrm{Fe}^{3+}\right)_{\operatorname{Red}}-\left(\mathrm{Ce}^{3+}\right)_{\operatorname{Red}} \\ &= 0.76-1.6 \\ & \mathrm{E}^{\mathrm{o}}=-0.84 \mathrm{~V} \end{aligned}$