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CHXII03:ELECTROCHEMISTRY

330206 How many electrons flow through the wire if a current of 1.5 ampere flow through it for 3hours?

1

1.60 \times 10^{19}

2

1.01 \times 10^{23}

3

1.01 \times 10^{19}

4

1.60 \times 10^{23}

Explanation:

$I = 1.5 A, t = 3 hr = 3 \times 60 \times 60 s$
$= 10800 s$
$Q = It = 1.5 \times 10800 = 16200 C$
and $1.602 \times 10^{- 19} C = 1 e^{-}$ (charge of one electron)
$∴ 16200 C = \frac{1 e^{-} \times 16200 C}{1.6 \times 10^{- 19} C}$
$= 10125 \times 10^{19} e^{-}$
$= 1.01 \times 10^{23} e^{-}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330207 Calculate mass of a divalent metal produced at cathode by passing $5 amp$ current through its salt solution for 100 minutes.
$( molar mass of metal = x)$

1

\frac{15}{193} x

2

\frac{193}{30} x

3

\frac{193}{30} x

4

\frac{193}{15} x

Explanation:

$t = 100 min = 100 \times 60 sec = 6000 sec$
Charge $=$ current $\times$ time
$= 5 amp \times 6000 sec = 30000 C$
According to the reaction, $M_{(aq)}^{2 +} + 2 \overset{―}{e} \to M_{(s)}$
$(∵$ Metal is divalent)
$2 F$ or $2 \times 96500 C$ is required to deposit 1 mole or ' $= x$ ' $= g$ of $M$
(Given, atomic mass of metal $= x$
$∴ F o r 30000 C$ , the mass of metal deposited
$= \frac{x \times 30000}{2 \times 96500} = \frac{x \times 300}{2 \times 965} = \frac{30x}{193}$

MHTCET - 2022

CHXII03:ELECTROCHEMISTRY

330208 The amount of current in faraday required for the reduction of 1 mol of $C r_{2} O_{7}^{2 -} i o n s t o C r^{3 +}$ is

1 1 F

2 2 F

3 6 F

4 4 F

Explanation:

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
6F of current required.

KCET - 2018

CHXII03:ELECTROCHEMISTRY

330209 An electrolytic cell contains a solution of ${Ag}_{2} {SO}_{4}$ and has platinum electrodes. A current is passed until 1.6 g of $O_{2}$ has been liberated at anode. The amount of silver deposited at cathode would be

1 107.88 g

2 1.6 g

3 0.8 g

4 21.60 g

Explanation:

$C a t h o d e : 2 A g^{+} + 2 e^{-} \to 2 A g$
$A n o d e : 2 O H^{-} - 2 e^{-} \to 2 O H$
$2 O H \to H_{2} O + \frac{1}{2} O_{2}$
$2 m o l e s o f e l e c t r o n s o r 2 \times 96500 C$
$l i b e r a t e s 16 g o f O_{2}$
$∴ 1.6 g o f O_{2} w i l l b e l i b e r a t e d b y$
$\frac{2 \times 96500}{16} \times 1.6 C = 19300 C$
$N o w, 2 \times 96500 C d e p o s i t s 2 \times 107.8 g o f A g$
$∴ 19300 C w i l l d e p o s i t$
$\frac{2 \times 107.8}{2 \times 96500} \times 19300 = 21.56 g A g$

CHXII03:ELECTROCHEMISTRY

330206 How many electrons flow through the wire if a current of 1.5 ampere flow through it for 3hours?

1

1.60 \times 10^{19}

2

1.01 \times 10^{23}

3

1.01 \times 10^{19}

4

1.60 \times 10^{23}

Explanation:

$I = 1.5 A, t = 3 hr = 3 \times 60 \times 60 s$
$= 10800 s$
$Q = It = 1.5 \times 10800 = 16200 C$
and $1.602 \times 10^{- 19} C = 1 e^{-}$ (charge of one electron)
$∴ 16200 C = \frac{1 e^{-} \times 16200 C}{1.6 \times 10^{- 19} C}$
$= 10125 \times 10^{19} e^{-}$
$= 1.01 \times 10^{23} e^{-}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330207 Calculate mass of a divalent metal produced at cathode by passing $5 amp$ current through its salt solution for 100 minutes.
$( molar mass of metal = x)$

1

\frac{15}{193} x

2

\frac{193}{30} x

3

\frac{193}{30} x

4

\frac{193}{15} x

Explanation:

$t = 100 min = 100 \times 60 sec = 6000 sec$
Charge $=$ current $\times$ time
$= 5 amp \times 6000 sec = 30000 C$
According to the reaction, $M_{(aq)}^{2 +} + 2 \overset{―}{e} \to M_{(s)}$
$(∵$ Metal is divalent)
$2 F$ or $2 \times 96500 C$ is required to deposit 1 mole or ' $= x$ ' $= g$ of $M$
(Given, atomic mass of metal $= x$
$∴ F o r 30000 C$ , the mass of metal deposited
$= \frac{x \times 30000}{2 \times 96500} = \frac{x \times 300}{2 \times 965} = \frac{30x}{193}$

MHTCET - 2022

CHXII03:ELECTROCHEMISTRY

330208 The amount of current in faraday required for the reduction of 1 mol of $C r_{2} O_{7}^{2 -} i o n s t o C r^{3 +}$ is

1 1 F

2 2 F

3 6 F

4 4 F

Explanation:

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
6F of current required.

KCET - 2018

CHXII03:ELECTROCHEMISTRY

330209 An electrolytic cell contains a solution of ${Ag}_{2} {SO}_{4}$ and has platinum electrodes. A current is passed until 1.6 g of $O_{2}$ has been liberated at anode. The amount of silver deposited at cathode would be

1 107.88 g

2 1.6 g

3 0.8 g

4 21.60 g

Explanation:

$C a t h o d e : 2 A g^{+} + 2 e^{-} \to 2 A g$
$A n o d e : 2 O H^{-} - 2 e^{-} \to 2 O H$
$2 O H \to H_{2} O + \frac{1}{2} O_{2}$
$2 m o l e s o f e l e c t r o n s o r 2 \times 96500 C$
$l i b e r a t e s 16 g o f O_{2}$
$∴ 1.6 g o f O_{2} w i l l b e l i b e r a t e d b y$
$\frac{2 \times 96500}{16} \times 1.6 C = 19300 C$
$N o w, 2 \times 96500 C d e p o s i t s 2 \times 107.8 g o f A g$
$∴ 19300 C w i l l d e p o s i t$
$\frac{2 \times 107.8}{2 \times 96500} \times 19300 = 21.56 g A g$

CHXII03:ELECTROCHEMISTRY

330206 How many electrons flow through the wire if a current of 1.5 ampere flow through it for 3hours?

1

1.60 \times 10^{19}

2

1.01 \times 10^{23}

3

1.01 \times 10^{19}

4

1.60 \times 10^{23}

Explanation:

$I = 1.5 A, t = 3 hr = 3 \times 60 \times 60 s$
$= 10800 s$
$Q = It = 1.5 \times 10800 = 16200 C$
and $1.602 \times 10^{- 19} C = 1 e^{-}$ (charge of one electron)
$∴ 16200 C = \frac{1 e^{-} \times 16200 C}{1.6 \times 10^{- 19} C}$
$= 10125 \times 10^{19} e^{-}$
$= 1.01 \times 10^{23} e^{-}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330207 Calculate mass of a divalent metal produced at cathode by passing $5 amp$ current through its salt solution for 100 minutes.
$( molar mass of metal = x)$

1

\frac{15}{193} x

2

\frac{193}{30} x

3

\frac{193}{30} x

4

\frac{193}{15} x

Explanation:

$t = 100 min = 100 \times 60 sec = 6000 sec$
Charge $=$ current $\times$ time
$= 5 amp \times 6000 sec = 30000 C$
According to the reaction, $M_{(aq)}^{2 +} + 2 \overset{―}{e} \to M_{(s)}$
$(∵$ Metal is divalent)
$2 F$ or $2 \times 96500 C$ is required to deposit 1 mole or ' $= x$ ' $= g$ of $M$
(Given, atomic mass of metal $= x$
$∴ F o r 30000 C$ , the mass of metal deposited
$= \frac{x \times 30000}{2 \times 96500} = \frac{x \times 300}{2 \times 965} = \frac{30x}{193}$

MHTCET - 2022

CHXII03:ELECTROCHEMISTRY

330208 The amount of current in faraday required for the reduction of 1 mol of $C r_{2} O_{7}^{2 -} i o n s t o C r^{3 +}$ is

1 1 F

2 2 F

3 6 F

4 4 F

Explanation:

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
6F of current required.

KCET - 2018

CHXII03:ELECTROCHEMISTRY

330209 An electrolytic cell contains a solution of ${Ag}_{2} {SO}_{4}$ and has platinum electrodes. A current is passed until 1.6 g of $O_{2}$ has been liberated at anode. The amount of silver deposited at cathode would be

1 107.88 g

2 1.6 g

3 0.8 g

4 21.60 g

Explanation:

$C a t h o d e : 2 A g^{+} + 2 e^{-} \to 2 A g$
$A n o d e : 2 O H^{-} - 2 e^{-} \to 2 O H$
$2 O H \to H_{2} O + \frac{1}{2} O_{2}$
$2 m o l e s o f e l e c t r o n s o r 2 \times 96500 C$
$l i b e r a t e s 16 g o f O_{2}$
$∴ 1.6 g o f O_{2} w i l l b e l i b e r a t e d b y$
$\frac{2 \times 96500}{16} \times 1.6 C = 19300 C$
$N o w, 2 \times 96500 C d e p o s i t s 2 \times 107.8 g o f A g$
$∴ 19300 C w i l l d e p o s i t$
$\frac{2 \times 107.8}{2 \times 96500} \times 19300 = 21.56 g A g$

CHXII03:ELECTROCHEMISTRY

330206 How many electrons flow through the wire if a current of 1.5 ampere flow through it for 3hours?

1

1.60 \times 10^{19}

2

1.01 \times 10^{23}

3

1.01 \times 10^{19}

4

1.60 \times 10^{23}

Explanation:

$I = 1.5 A, t = 3 hr = 3 \times 60 \times 60 s$
$= 10800 s$
$Q = It = 1.5 \times 10800 = 16200 C$
and $1.602 \times 10^{- 19} C = 1 e^{-}$ (charge of one electron)
$∴ 16200 C = \frac{1 e^{-} \times 16200 C}{1.6 \times 10^{- 19} C}$
$= 10125 \times 10^{19} e^{-}$
$= 1.01 \times 10^{23} e^{-}$

MHTCET - 2021

CHXII03:ELECTROCHEMISTRY

330207 Calculate mass of a divalent metal produced at cathode by passing $5 amp$ current through its salt solution for 100 minutes.
$( molar mass of metal = x)$

1

\frac{15}{193} x

2

\frac{193}{30} x

3

\frac{193}{30} x

4

\frac{193}{15} x

Explanation:

$t = 100 min = 100 \times 60 sec = 6000 sec$
Charge $=$ current $\times$ time
$= 5 amp \times 6000 sec = 30000 C$
According to the reaction, $M_{(aq)}^{2 +} + 2 \overset{―}{e} \to M_{(s)}$
$(∵$ Metal is divalent)
$2 F$ or $2 \times 96500 C$ is required to deposit 1 mole or ' $= x$ ' $= g$ of $M$
(Given, atomic mass of metal $= x$
$∴ F o r 30000 C$ , the mass of metal deposited
$= \frac{x \times 30000}{2 \times 96500} = \frac{x \times 300}{2 \times 965} = \frac{30x}{193}$

MHTCET - 2022

CHXII03:ELECTROCHEMISTRY

330208 The amount of current in faraday required for the reduction of 1 mol of $C r_{2} O_{7}^{2 -} i o n s t o C r^{3 +}$ is

1 1 F

2 2 F

3 6 F

4 4 F

Explanation:

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$
6F of current required.

KCET - 2018

CHXII03:ELECTROCHEMISTRY

330209 An electrolytic cell contains a solution of ${Ag}_{2} {SO}_{4}$ and has platinum electrodes. A current is passed until 1.6 g of $O_{2}$ has been liberated at anode. The amount of silver deposited at cathode would be

1 107.88 g

2 1.6 g

3 0.8 g

4 21.60 g

Explanation:

$C a t h o d e : 2 A g^{+} + 2 e^{-} \to 2 A g$
$A n o d e : 2 O H^{-} - 2 e^{-} \to 2 O H$
$2 O H \to H_{2} O + \frac{1}{2} O_{2}$
$2 m o l e s o f e l e c t r o n s o r 2 \times 96500 C$
$l i b e r a t e s 16 g o f O_{2}$
$∴ 1.6 g o f O_{2} w i l l b e l i b e r a t e d b y$
$\frac{2 \times 96500}{16} \times 1.6 C = 19300 C$
$N o w, 2 \times 96500 C d e p o s i t s 2 \times 107.8 g o f A g$
$∴ 19300 C w i l l d e p o s i t$
$\frac{2 \times 107.8}{2 \times 96500} \times 19300 = 21.56 g A g$