QBManager

ELECTROCHEMISTRY

276035 A $100.0 mL$ dilute solution of ${Ag}^{+}$ is electrolysed for $15.0$ minutes with a current of $1.25 mA$ and the silver is removed completely. What was the initial $[{Ag}^{+}]$ ?

1

2.32 \times 10^{- 1}

2

2.32 \times 10^{- 4}

3

2.32 \times 10^{- 3}

4

1.16 \times 10^{- 4}

Explanation:

Given, $V = 100 mL, t = 15 \min, I = 1.25 mA$ $Q = I \times t = 1.25 \times 10^{- 3} \times 15 \times 60 = 1.125 C$
Cathodic reaction,
$\begin{aligned} {Ag}^{+} + e^{-} \to Ag \\ 1 F \to 108 g \\ ∴ 96500 C \to 108 g \\ 1.125 C \to \frac{108 \times 1.125}{96500} \\ = 1.259 \times 10^{- 3} g \\ Conc. [{Ag}^{+}] = \frac{1.259 \times 10^{- 3} \times 10^{3}}{108 \times 100} = 1.16 \times 10^{- 4} \end{aligned}$

BITSAT - 2018

ELECTROCHEMISTRY

276036 One coulomb of charge passes through a solution of ${AgNO}_{3}$ and ${CuSO}_{4}$ connected in series and the concentration of two solutions being in the ratio $1 : 2$ . The ratio of amount of $Ag$ and $Cu$ deposited on Pt electrode is

1 107.9:63.54

2

54 : 31.77

3

107.9 : 31.77

4

54 : 63.54

Explanation:

${Ag}^{+} + 1 e^{-} \to Ag (s)$
${Cu}^{2 +} + 2 e^{-} \to Cu (s)$
As Faraday's ${II}^{nd}$ law states that ratio of the weight deposited on the electron is equal to the ratio of the equivalent weight of the substance.
$Hence, \begin{aligned} \frac{w_{1}}{w_{2}} = \frac{E_{1}}{E_{2}} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{E_{Ag}}{E_{Cu}} \\ \frac{107.9}{1} \times \frac{2}{63.54} = \frac{107.9}{31.77} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{107.9}{31.77} \end{aligned}$

CG PET- 2016

ELECTROCHEMISTRY

276037 How much Coulomb needed to convert 1 mole of ${MnO}_{4}$ into ${Mn}^{2 +}$ ?

1

482500 C

2

193000 C

3

96500 C

4

36500 C

Explanation:

${MnO}_{4}^{-} + 5 e^{-} \to {Mn}^{2 +}$
Here, 5 moles of electrons are needed for reduction of 1 mole of ${MnO}_{4}^{-}$ to ${Mn}^{2 +}$
5 moles of electron $= 5$ Faraday
$\Rightarrow$ Quantity of charge needed $= 5 \times 96500$
$= 482500 C$

CG PET-2013

ELECTROCHEMISTRY

276038 $4.5 g$ of aluminum (atomic mass $27 amu$ ) is deposited at cathode from ${Al}^{3 +}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $H^{+}$ ions in solution by the same quantity of electric charge will be

1

44.8 L

2

11.2 L

3

22.4 L

4

5.6 L

Explanation:

$m = \frac{eq. wt \times Q}{96485}$
$\begin{aligned} \frac{4.5 \times 96485}{27 / 3} = \frac{m \times 96485}{1} \\ \frac{4.5 \times 3}{27} = m \\ m = 0.5 \end{aligned}$
Moles of $H_{2} = \frac{0.5}{2} = 0.25$
Volume $= 0.25 \times 22.4$
$= 5.6 L$

CG PET-2007

ELECTROCHEMISTRY

276039 The weight of $Cu$ deposited, when 2 Faraday of electricity is passed will be

1

31.75 g

2

23.85 g

3

63.5 g

4

125.67 g

Explanation:

$\begin{aligned} {Cu}^{+ 2} + 2 e^{-} ⟶ Cu \\ 2 F = 1 mole Cu \\ = 63.5 gm \end{aligned}$

CG PET-2018

ELECTROCHEMISTRY

276035 A $100.0 mL$ dilute solution of ${Ag}^{+}$ is electrolysed for $15.0$ minutes with a current of $1.25 mA$ and the silver is removed completely. What was the initial $[{Ag}^{+}]$ ?

1

2.32 \times 10^{- 1}

2

2.32 \times 10^{- 4}

3

2.32 \times 10^{- 3}

4

1.16 \times 10^{- 4}

Explanation:

Given, $V = 100 mL, t = 15 \min, I = 1.25 mA$ $Q = I \times t = 1.25 \times 10^{- 3} \times 15 \times 60 = 1.125 C$
Cathodic reaction,
$\begin{aligned} {Ag}^{+} + e^{-} \to Ag \\ 1 F \to 108 g \\ ∴ 96500 C \to 108 g \\ 1.125 C \to \frac{108 \times 1.125}{96500} \\ = 1.259 \times 10^{- 3} g \\ Conc. [{Ag}^{+}] = \frac{1.259 \times 10^{- 3} \times 10^{3}}{108 \times 100} = 1.16 \times 10^{- 4} \end{aligned}$

BITSAT - 2018

ELECTROCHEMISTRY

276036 One coulomb of charge passes through a solution of ${AgNO}_{3}$ and ${CuSO}_{4}$ connected in series and the concentration of two solutions being in the ratio $1 : 2$ . The ratio of amount of $Ag$ and $Cu$ deposited on Pt electrode is

1 107.9:63.54

2

54 : 31.77

3

107.9 : 31.77

4

54 : 63.54

Explanation:

${Ag}^{+} + 1 e^{-} \to Ag (s)$
${Cu}^{2 +} + 2 e^{-} \to Cu (s)$
As Faraday's ${II}^{nd}$ law states that ratio of the weight deposited on the electron is equal to the ratio of the equivalent weight of the substance.
$Hence, \begin{aligned} \frac{w_{1}}{w_{2}} = \frac{E_{1}}{E_{2}} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{E_{Ag}}{E_{Cu}} \\ \frac{107.9}{1} \times \frac{2}{63.54} = \frac{107.9}{31.77} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{107.9}{31.77} \end{aligned}$

CG PET- 2016

ELECTROCHEMISTRY

276037 How much Coulomb needed to convert 1 mole of ${MnO}_{4}$ into ${Mn}^{2 +}$ ?

1

482500 C

2

193000 C

3

96500 C

4

36500 C

Explanation:

${MnO}_{4}^{-} + 5 e^{-} \to {Mn}^{2 +}$
Here, 5 moles of electrons are needed for reduction of 1 mole of ${MnO}_{4}^{-}$ to ${Mn}^{2 +}$
5 moles of electron $= 5$ Faraday
$\Rightarrow$ Quantity of charge needed $= 5 \times 96500$
$= 482500 C$

CG PET-2013

ELECTROCHEMISTRY

276038 $4.5 g$ of aluminum (atomic mass $27 amu$ ) is deposited at cathode from ${Al}^{3 +}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $H^{+}$ ions in solution by the same quantity of electric charge will be

1

44.8 L

2

11.2 L

3

22.4 L

4

5.6 L

Explanation:

$m = \frac{eq. wt \times Q}{96485}$
$\begin{aligned} \frac{4.5 \times 96485}{27 / 3} = \frac{m \times 96485}{1} \\ \frac{4.5 \times 3}{27} = m \\ m = 0.5 \end{aligned}$
Moles of $H_{2} = \frac{0.5}{2} = 0.25$
Volume $= 0.25 \times 22.4$
$= 5.6 L$

CG PET-2007

ELECTROCHEMISTRY

276039 The weight of $Cu$ deposited, when 2 Faraday of electricity is passed will be

1

31.75 g

2

23.85 g

3

63.5 g

4

125.67 g

Explanation:

$\begin{aligned} {Cu}^{+ 2} + 2 e^{-} ⟶ Cu \\ 2 F = 1 mole Cu \\ = 63.5 gm \end{aligned}$

CG PET-2018

ELECTROCHEMISTRY

276035 A $100.0 mL$ dilute solution of ${Ag}^{+}$ is electrolysed for $15.0$ minutes with a current of $1.25 mA$ and the silver is removed completely. What was the initial $[{Ag}^{+}]$ ?

1

2.32 \times 10^{- 1}

2

2.32 \times 10^{- 4}

3

2.32 \times 10^{- 3}

4

1.16 \times 10^{- 4}

Explanation:

Given, $V = 100 mL, t = 15 \min, I = 1.25 mA$ $Q = I \times t = 1.25 \times 10^{- 3} \times 15 \times 60 = 1.125 C$
Cathodic reaction,
$\begin{aligned} {Ag}^{+} + e^{-} \to Ag \\ 1 F \to 108 g \\ ∴ 96500 C \to 108 g \\ 1.125 C \to \frac{108 \times 1.125}{96500} \\ = 1.259 \times 10^{- 3} g \\ Conc. [{Ag}^{+}] = \frac{1.259 \times 10^{- 3} \times 10^{3}}{108 \times 100} = 1.16 \times 10^{- 4} \end{aligned}$

BITSAT - 2018

ELECTROCHEMISTRY

276036 One coulomb of charge passes through a solution of ${AgNO}_{3}$ and ${CuSO}_{4}$ connected in series and the concentration of two solutions being in the ratio $1 : 2$ . The ratio of amount of $Ag$ and $Cu$ deposited on Pt electrode is

1 107.9:63.54

2

54 : 31.77

3

107.9 : 31.77

4

54 : 63.54

Explanation:

${Ag}^{+} + 1 e^{-} \to Ag (s)$
${Cu}^{2 +} + 2 e^{-} \to Cu (s)$
As Faraday's ${II}^{nd}$ law states that ratio of the weight deposited on the electron is equal to the ratio of the equivalent weight of the substance.
$Hence, \begin{aligned} \frac{w_{1}}{w_{2}} = \frac{E_{1}}{E_{2}} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{E_{Ag}}{E_{Cu}} \\ \frac{107.9}{1} \times \frac{2}{63.54} = \frac{107.9}{31.77} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{107.9}{31.77} \end{aligned}$

CG PET- 2016

ELECTROCHEMISTRY

276037 How much Coulomb needed to convert 1 mole of ${MnO}_{4}$ into ${Mn}^{2 +}$ ?

1

482500 C

2

193000 C

3

96500 C

4

36500 C

Explanation:

${MnO}_{4}^{-} + 5 e^{-} \to {Mn}^{2 +}$
Here, 5 moles of electrons are needed for reduction of 1 mole of ${MnO}_{4}^{-}$ to ${Mn}^{2 +}$
5 moles of electron $= 5$ Faraday
$\Rightarrow$ Quantity of charge needed $= 5 \times 96500$
$= 482500 C$

CG PET-2013

ELECTROCHEMISTRY

276038 $4.5 g$ of aluminum (atomic mass $27 amu$ ) is deposited at cathode from ${Al}^{3 +}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $H^{+}$ ions in solution by the same quantity of electric charge will be

1

44.8 L

2

11.2 L

3

22.4 L

4

5.6 L

Explanation:

$m = \frac{eq. wt \times Q}{96485}$
$\begin{aligned} \frac{4.5 \times 96485}{27 / 3} = \frac{m \times 96485}{1} \\ \frac{4.5 \times 3}{27} = m \\ m = 0.5 \end{aligned}$
Moles of $H_{2} = \frac{0.5}{2} = 0.25$
Volume $= 0.25 \times 22.4$
$= 5.6 L$

CG PET-2007

ELECTROCHEMISTRY

276039 The weight of $Cu$ deposited, when 2 Faraday of electricity is passed will be

1

31.75 g

2

23.85 g

3

63.5 g

4

125.67 g

Explanation:

$\begin{aligned} {Cu}^{+ 2} + 2 e^{-} ⟶ Cu \\ 2 F = 1 mole Cu \\ = 63.5 gm \end{aligned}$

CG PET-2018

ELECTROCHEMISTRY

276035 A $100.0 mL$ dilute solution of ${Ag}^{+}$ is electrolysed for $15.0$ minutes with a current of $1.25 mA$ and the silver is removed completely. What was the initial $[{Ag}^{+}]$ ?

1

2.32 \times 10^{- 1}

2

2.32 \times 10^{- 4}

3

2.32 \times 10^{- 3}

4

1.16 \times 10^{- 4}

Explanation:

Given, $V = 100 mL, t = 15 \min, I = 1.25 mA$ $Q = I \times t = 1.25 \times 10^{- 3} \times 15 \times 60 = 1.125 C$
Cathodic reaction,
$\begin{aligned} {Ag}^{+} + e^{-} \to Ag \\ 1 F \to 108 g \\ ∴ 96500 C \to 108 g \\ 1.125 C \to \frac{108 \times 1.125}{96500} \\ = 1.259 \times 10^{- 3} g \\ Conc. [{Ag}^{+}] = \frac{1.259 \times 10^{- 3} \times 10^{3}}{108 \times 100} = 1.16 \times 10^{- 4} \end{aligned}$

BITSAT - 2018

ELECTROCHEMISTRY

276036 One coulomb of charge passes through a solution of ${AgNO}_{3}$ and ${CuSO}_{4}$ connected in series and the concentration of two solutions being in the ratio $1 : 2$ . The ratio of amount of $Ag$ and $Cu$ deposited on Pt electrode is

1 107.9:63.54

2

54 : 31.77

3

107.9 : 31.77

4

54 : 63.54

Explanation:

${Ag}^{+} + 1 e^{-} \to Ag (s)$
${Cu}^{2 +} + 2 e^{-} \to Cu (s)$
As Faraday's ${II}^{nd}$ law states that ratio of the weight deposited on the electron is equal to the ratio of the equivalent weight of the substance.
$Hence, \begin{aligned} \frac{w_{1}}{w_{2}} = \frac{E_{1}}{E_{2}} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{E_{Ag}}{E_{Cu}} \\ \frac{107.9}{1} \times \frac{2}{63.54} = \frac{107.9}{31.77} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{107.9}{31.77} \end{aligned}$

CG PET- 2016

ELECTROCHEMISTRY

276037 How much Coulomb needed to convert 1 mole of ${MnO}_{4}$ into ${Mn}^{2 +}$ ?

1

482500 C

2

193000 C

3

96500 C

4

36500 C

Explanation:

${MnO}_{4}^{-} + 5 e^{-} \to {Mn}^{2 +}$
Here, 5 moles of electrons are needed for reduction of 1 mole of ${MnO}_{4}^{-}$ to ${Mn}^{2 +}$
5 moles of electron $= 5$ Faraday
$\Rightarrow$ Quantity of charge needed $= 5 \times 96500$
$= 482500 C$

CG PET-2013

ELECTROCHEMISTRY

276038 $4.5 g$ of aluminum (atomic mass $27 amu$ ) is deposited at cathode from ${Al}^{3 +}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $H^{+}$ ions in solution by the same quantity of electric charge will be

1

44.8 L

2

11.2 L

3

22.4 L

4

5.6 L

Explanation:

$m = \frac{eq. wt \times Q}{96485}$
$\begin{aligned} \frac{4.5 \times 96485}{27 / 3} = \frac{m \times 96485}{1} \\ \frac{4.5 \times 3}{27} = m \\ m = 0.5 \end{aligned}$
Moles of $H_{2} = \frac{0.5}{2} = 0.25$
Volume $= 0.25 \times 22.4$
$= 5.6 L$

CG PET-2007

ELECTROCHEMISTRY

276039 The weight of $Cu$ deposited, when 2 Faraday of electricity is passed will be

1

31.75 g

2

23.85 g

3

63.5 g

4

125.67 g

Explanation:

$\begin{aligned} {Cu}^{+ 2} + 2 e^{-} ⟶ Cu \\ 2 F = 1 mole Cu \\ = 63.5 gm \end{aligned}$

CG PET-2018

ELECTROCHEMISTRY

276035 A $100.0 mL$ dilute solution of ${Ag}^{+}$ is electrolysed for $15.0$ minutes with a current of $1.25 mA$ and the silver is removed completely. What was the initial $[{Ag}^{+}]$ ?

1

2.32 \times 10^{- 1}

2

2.32 \times 10^{- 4}

3

2.32 \times 10^{- 3}

4

1.16 \times 10^{- 4}

Explanation:

Given, $V = 100 mL, t = 15 \min, I = 1.25 mA$ $Q = I \times t = 1.25 \times 10^{- 3} \times 15 \times 60 = 1.125 C$
Cathodic reaction,
$\begin{aligned} {Ag}^{+} + e^{-} \to Ag \\ 1 F \to 108 g \\ ∴ 96500 C \to 108 g \\ 1.125 C \to \frac{108 \times 1.125}{96500} \\ = 1.259 \times 10^{- 3} g \\ Conc. [{Ag}^{+}] = \frac{1.259 \times 10^{- 3} \times 10^{3}}{108 \times 100} = 1.16 \times 10^{- 4} \end{aligned}$

BITSAT - 2018

ELECTROCHEMISTRY

276036 One coulomb of charge passes through a solution of ${AgNO}_{3}$ and ${CuSO}_{4}$ connected in series and the concentration of two solutions being in the ratio $1 : 2$ . The ratio of amount of $Ag$ and $Cu$ deposited on Pt electrode is

1 107.9:63.54

2

54 : 31.77

3

107.9 : 31.77

4

54 : 63.54

Explanation:

${Ag}^{+} + 1 e^{-} \to Ag (s)$
${Cu}^{2 +} + 2 e^{-} \to Cu (s)$
As Faraday's ${II}^{nd}$ law states that ratio of the weight deposited on the electron is equal to the ratio of the equivalent weight of the substance.
$Hence, \begin{aligned} \frac{w_{1}}{w_{2}} = \frac{E_{1}}{E_{2}} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{E_{Ag}}{E_{Cu}} \\ \frac{107.9}{1} \times \frac{2}{63.54} = \frac{107.9}{31.77} \\ \frac{w_{Ag}}{w_{Cu}} = \frac{107.9}{31.77} \end{aligned}$

CG PET- 2016

ELECTROCHEMISTRY

276037 How much Coulomb needed to convert 1 mole of ${MnO}_{4}$ into ${Mn}^{2 +}$ ?

1

482500 C

2

193000 C

3

96500 C

4

36500 C

Explanation:

${MnO}_{4}^{-} + 5 e^{-} \to {Mn}^{2 +}$
Here, 5 moles of electrons are needed for reduction of 1 mole of ${MnO}_{4}^{-}$ to ${Mn}^{2 +}$
5 moles of electron $= 5$ Faraday
$\Rightarrow$ Quantity of charge needed $= 5 \times 96500$
$= 482500 C$

CG PET-2013

ELECTROCHEMISTRY

276038 $4.5 g$ of aluminum (atomic mass $27 amu$ ) is deposited at cathode from ${Al}^{3 +}$ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from $H^{+}$ ions in solution by the same quantity of electric charge will be

1

44.8 L

2

11.2 L

3

22.4 L

4

5.6 L

Explanation:

$m = \frac{eq. wt \times Q}{96485}$
$\begin{aligned} \frac{4.5 \times 96485}{27 / 3} = \frac{m \times 96485}{1} \\ \frac{4.5 \times 3}{27} = m \\ m = 0.5 \end{aligned}$
Moles of $H_{2} = \frac{0.5}{2} = 0.25$
Volume $= 0.25 \times 22.4$
$= 5.6 L$

CG PET-2007

ELECTROCHEMISTRY

276039 The weight of $Cu$ deposited, when 2 Faraday of electricity is passed will be

1

31.75 g

2

23.85 g

3

63.5 g

4

125.67 g

Explanation:

$\begin{aligned} {Cu}^{+ 2} + 2 e^{-} ⟶ Cu \\ 2 F = 1 mole Cu \\ = 63.5 gm \end{aligned}$

CG PET-2018