QBManager

ELECTROCHEMISTRY

276252 In the lead-acid battery during charging, the cathode reaction is

1 formation of

{PbO}_{2}

2 formation of

{PbSO}_{4}

3 reduction of

{Pb}^{2 +}

to

Pb

4 decomposition of

Pb

at the anode.

Explanation:

During charging of lead-acid battery ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on cathode.
At
Anode
${PbSO}_{4} (s) + 2 H_{2} O \to {PbO}_{2} (s) + {SO}_{4}^{2 -} (aq) + 4 H^{+} (aq) + 2 e^{-}$
At Cathode ${PbSO}_{4} (s) + 2 e^{-} \to Pb (s) + {SO}_{4}^{2 -} (aq)$
Thus, ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on the cathode while ${PbSO}_{4}$ is oxidized to ${PbO}_{2}$ at anode.

AMU-2015

ELECTROCHEMISTRY

276257 The electrochemical cell stops working after sometime because :

1 electrode potential of both the electrodes becomes zero

2 electrode potential of both the electrodes becomes equal

3 one of the electrodes is eaten away

4 the cell reaction gets reversed

Explanation:

As we know that,
$Δ E_{cell} = E_{cathode} - E_{anode}$
when $E_{cathode} = E_{anode}$
$Δ E_{cell} = 0$
If $Δ E_{cell} = 0$
When electrode potential of both the electrodes becomes equal then the electrochemical cell stops working after sometime.

VITEEE- 2008

ELECTROCHEMISTRY

276258 The extent of charge of lead accumulator is determined by

1 amount of

{PbSO}_{4}

in the battery

2 amount of

{PbO}_{2}

in the battery

3 specific gravity of

H_{2} {SO}_{4}

of the battery

4 amount of

Pb

in the battery

Explanation:

The lead accumulator refers to a secondary cell as the electrical energy does not generate itself inside the cell, but it is stored prior to an external source.
The extent of charge of a lead accumulator is determined by the specific gravity of $H_{2} {SO}_{4}$ of the battery.
$\begin{matrix} Pb (s) + {PbO}_{2} (s) + 4 H^{+} (aq) + 2 {SO}_{4}^{2 -} (aq) ⟶ \\ 2 {PbSO}_{4} (l) + 2 H_{2} O \end{matrix}$
The fully charged lead accumulator is $1.30 g / ml$ and when it drops below $1.20 g / ml$ , the lead accumulator need recharging.

AP-EAMCET (Engg.) 2013

ELECTROCHEMISTRY

276262 Other things being equal, the EMF of a Daniel cell may be increased by

1 keeping low temperature

2 using large copper electrodes

3 using large zinc electrodes

4 decreasing concentration of

{Cu}^{2 +}

ions

Explanation:

For Daniel cell
$Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu$
$E_{cell} = E_{cell}^{o} - \frac{RT}{nF} \ln \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}$
The decrease in temperature and increase in $[{Cu}^{2 +}]$ will increase the EMF.
Thus, low temperature will increases EMF of Daniel cell.

BCECE-2011

ELECTROCHEMISTRY

276264 In a Daniell cell constructed in the laboratory, the voltage observed was $0.9 V$ instead of $1.1 V$ of the standard cell. A possible explanation is-

1

[{Zn}^{2 +}] > [{Cu}^{2 +}]

2

[{Zn}^{2 +}] < [{Cu}^{2 +}]

3

Zn

electrode has twice the surface of

Cu

electrode

4 mol ratio of

{Zn}^{2 +} : {Cu}^{2 +}

is

2 : 1

Explanation:

$\begin{aligned} E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 1.1 - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ E_{cell} < 1.1 V if \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} > 1 \end{aligned}$
Thus, option (a) is correct.

BCECE-2013

ELECTROCHEMISTRY

276252 In the lead-acid battery during charging, the cathode reaction is

1 formation of

{PbO}_{2}

2 formation of

{PbSO}_{4}

3 reduction of

{Pb}^{2 +}

to

Pb

4 decomposition of

Pb

at the anode.

Explanation:

During charging of lead-acid battery ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on cathode.
At
Anode
${PbSO}_{4} (s) + 2 H_{2} O \to {PbO}_{2} (s) + {SO}_{4}^{2 -} (aq) + 4 H^{+} (aq) + 2 e^{-}$
At Cathode ${PbSO}_{4} (s) + 2 e^{-} \to Pb (s) + {SO}_{4}^{2 -} (aq)$
Thus, ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on the cathode while ${PbSO}_{4}$ is oxidized to ${PbO}_{2}$ at anode.

AMU-2015

ELECTROCHEMISTRY

276257 The electrochemical cell stops working after sometime because :

1 electrode potential of both the electrodes becomes zero

2 electrode potential of both the electrodes becomes equal

3 one of the electrodes is eaten away

4 the cell reaction gets reversed

Explanation:

As we know that,
$Δ E_{cell} = E_{cathode} - E_{anode}$
when $E_{cathode} = E_{anode}$
$Δ E_{cell} = 0$
If $Δ E_{cell} = 0$
When electrode potential of both the electrodes becomes equal then the electrochemical cell stops working after sometime.

VITEEE- 2008

ELECTROCHEMISTRY

276258 The extent of charge of lead accumulator is determined by

1 amount of

{PbSO}_{4}

in the battery

2 amount of

{PbO}_{2}

in the battery

3 specific gravity of

H_{2} {SO}_{4}

of the battery

4 amount of

Pb

in the battery

Explanation:

The lead accumulator refers to a secondary cell as the electrical energy does not generate itself inside the cell, but it is stored prior to an external source.
The extent of charge of a lead accumulator is determined by the specific gravity of $H_{2} {SO}_{4}$ of the battery.
$\begin{matrix} Pb (s) + {PbO}_{2} (s) + 4 H^{+} (aq) + 2 {SO}_{4}^{2 -} (aq) ⟶ \\ 2 {PbSO}_{4} (l) + 2 H_{2} O \end{matrix}$
The fully charged lead accumulator is $1.30 g / ml$ and when it drops below $1.20 g / ml$ , the lead accumulator need recharging.

AP-EAMCET (Engg.) 2013

ELECTROCHEMISTRY

276262 Other things being equal, the EMF of a Daniel cell may be increased by

1 keeping low temperature

2 using large copper electrodes

3 using large zinc electrodes

4 decreasing concentration of

{Cu}^{2 +}

ions

Explanation:

For Daniel cell
$Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu$
$E_{cell} = E_{cell}^{o} - \frac{RT}{nF} \ln \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}$
The decrease in temperature and increase in $[{Cu}^{2 +}]$ will increase the EMF.
Thus, low temperature will increases EMF of Daniel cell.

BCECE-2011

ELECTROCHEMISTRY

276264 In a Daniell cell constructed in the laboratory, the voltage observed was $0.9 V$ instead of $1.1 V$ of the standard cell. A possible explanation is-

1

[{Zn}^{2 +}] > [{Cu}^{2 +}]

2

[{Zn}^{2 +}] < [{Cu}^{2 +}]

3

Zn

electrode has twice the surface of

Cu

electrode

4 mol ratio of

{Zn}^{2 +} : {Cu}^{2 +}

is

2 : 1

Explanation:

$\begin{aligned} E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 1.1 - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ E_{cell} < 1.1 V if \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} > 1 \end{aligned}$
Thus, option (a) is correct.

BCECE-2013

ELECTROCHEMISTRY

276252 In the lead-acid battery during charging, the cathode reaction is

1 formation of

{PbO}_{2}

2 formation of

{PbSO}_{4}

3 reduction of

{Pb}^{2 +}

to

Pb

4 decomposition of

Pb

at the anode.

Explanation:

During charging of lead-acid battery ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on cathode.
At
Anode
${PbSO}_{4} (s) + 2 H_{2} O \to {PbO}_{2} (s) + {SO}_{4}^{2 -} (aq) + 4 H^{+} (aq) + 2 e^{-}$
At Cathode ${PbSO}_{4} (s) + 2 e^{-} \to Pb (s) + {SO}_{4}^{2 -} (aq)$
Thus, ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on the cathode while ${PbSO}_{4}$ is oxidized to ${PbO}_{2}$ at anode.

AMU-2015

ELECTROCHEMISTRY

276257 The electrochemical cell stops working after sometime because :

1 electrode potential of both the electrodes becomes zero

2 electrode potential of both the electrodes becomes equal

3 one of the electrodes is eaten away

4 the cell reaction gets reversed

Explanation:

As we know that,
$Δ E_{cell} = E_{cathode} - E_{anode}$
when $E_{cathode} = E_{anode}$
$Δ E_{cell} = 0$
If $Δ E_{cell} = 0$
When electrode potential of both the electrodes becomes equal then the electrochemical cell stops working after sometime.

VITEEE- 2008

ELECTROCHEMISTRY

276258 The extent of charge of lead accumulator is determined by

1 amount of

{PbSO}_{4}

in the battery

2 amount of

{PbO}_{2}

in the battery

3 specific gravity of

H_{2} {SO}_{4}

of the battery

4 amount of

Pb

in the battery

Explanation:

The lead accumulator refers to a secondary cell as the electrical energy does not generate itself inside the cell, but it is stored prior to an external source.
The extent of charge of a lead accumulator is determined by the specific gravity of $H_{2} {SO}_{4}$ of the battery.
$\begin{matrix} Pb (s) + {PbO}_{2} (s) + 4 H^{+} (aq) + 2 {SO}_{4}^{2 -} (aq) ⟶ \\ 2 {PbSO}_{4} (l) + 2 H_{2} O \end{matrix}$
The fully charged lead accumulator is $1.30 g / ml$ and when it drops below $1.20 g / ml$ , the lead accumulator need recharging.

AP-EAMCET (Engg.) 2013

ELECTROCHEMISTRY

276262 Other things being equal, the EMF of a Daniel cell may be increased by

1 keeping low temperature

2 using large copper electrodes

3 using large zinc electrodes

4 decreasing concentration of

{Cu}^{2 +}

ions

Explanation:

For Daniel cell
$Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu$
$E_{cell} = E_{cell}^{o} - \frac{RT}{nF} \ln \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}$
The decrease in temperature and increase in $[{Cu}^{2 +}]$ will increase the EMF.
Thus, low temperature will increases EMF of Daniel cell.

BCECE-2011

ELECTROCHEMISTRY

276264 In a Daniell cell constructed in the laboratory, the voltage observed was $0.9 V$ instead of $1.1 V$ of the standard cell. A possible explanation is-

1

[{Zn}^{2 +}] > [{Cu}^{2 +}]

2

[{Zn}^{2 +}] < [{Cu}^{2 +}]

3

Zn

electrode has twice the surface of

Cu

electrode

4 mol ratio of

{Zn}^{2 +} : {Cu}^{2 +}

is

2 : 1

Explanation:

$\begin{aligned} E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 1.1 - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ E_{cell} < 1.1 V if \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} > 1 \end{aligned}$
Thus, option (a) is correct.

BCECE-2013

ELECTROCHEMISTRY

276252 In the lead-acid battery during charging, the cathode reaction is

1 formation of

{PbO}_{2}

2 formation of

{PbSO}_{4}

3 reduction of

{Pb}^{2 +}

to

Pb

4 decomposition of

Pb

at the anode.

Explanation:

During charging of lead-acid battery ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on cathode.
At
Anode
${PbSO}_{4} (s) + 2 H_{2} O \to {PbO}_{2} (s) + {SO}_{4}^{2 -} (aq) + 4 H^{+} (aq) + 2 e^{-}$
At Cathode ${PbSO}_{4} (s) + 2 e^{-} \to Pb (s) + {SO}_{4}^{2 -} (aq)$
Thus, ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on the cathode while ${PbSO}_{4}$ is oxidized to ${PbO}_{2}$ at anode.

AMU-2015

ELECTROCHEMISTRY

276257 The electrochemical cell stops working after sometime because :

1 electrode potential of both the electrodes becomes zero

2 electrode potential of both the electrodes becomes equal

3 one of the electrodes is eaten away

4 the cell reaction gets reversed

Explanation:

As we know that,
$Δ E_{cell} = E_{cathode} - E_{anode}$
when $E_{cathode} = E_{anode}$
$Δ E_{cell} = 0$
If $Δ E_{cell} = 0$
When electrode potential of both the electrodes becomes equal then the electrochemical cell stops working after sometime.

VITEEE- 2008

ELECTROCHEMISTRY

276258 The extent of charge of lead accumulator is determined by

1 amount of

{PbSO}_{4}

in the battery

2 amount of

{PbO}_{2}

in the battery

3 specific gravity of

H_{2} {SO}_{4}

of the battery

4 amount of

Pb

in the battery

Explanation:

The lead accumulator refers to a secondary cell as the electrical energy does not generate itself inside the cell, but it is stored prior to an external source.
The extent of charge of a lead accumulator is determined by the specific gravity of $H_{2} {SO}_{4}$ of the battery.
$\begin{matrix} Pb (s) + {PbO}_{2} (s) + 4 H^{+} (aq) + 2 {SO}_{4}^{2 -} (aq) ⟶ \\ 2 {PbSO}_{4} (l) + 2 H_{2} O \end{matrix}$
The fully charged lead accumulator is $1.30 g / ml$ and when it drops below $1.20 g / ml$ , the lead accumulator need recharging.

AP-EAMCET (Engg.) 2013

ELECTROCHEMISTRY

276262 Other things being equal, the EMF of a Daniel cell may be increased by

1 keeping low temperature

2 using large copper electrodes

3 using large zinc electrodes

4 decreasing concentration of

{Cu}^{2 +}

ions

Explanation:

For Daniel cell
$Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu$
$E_{cell} = E_{cell}^{o} - \frac{RT}{nF} \ln \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}$
The decrease in temperature and increase in $[{Cu}^{2 +}]$ will increase the EMF.
Thus, low temperature will increases EMF of Daniel cell.

BCECE-2011

ELECTROCHEMISTRY

276264 In a Daniell cell constructed in the laboratory, the voltage observed was $0.9 V$ instead of $1.1 V$ of the standard cell. A possible explanation is-

1

[{Zn}^{2 +}] > [{Cu}^{2 +}]

2

[{Zn}^{2 +}] < [{Cu}^{2 +}]

3

Zn

electrode has twice the surface of

Cu

electrode

4 mol ratio of

{Zn}^{2 +} : {Cu}^{2 +}

is

2 : 1

Explanation:

$\begin{aligned} E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 1.1 - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ E_{cell} < 1.1 V if \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} > 1 \end{aligned}$
Thus, option (a) is correct.

BCECE-2013

ELECTROCHEMISTRY

276252 In the lead-acid battery during charging, the cathode reaction is

1 formation of

{PbO}_{2}

2 formation of

{PbSO}_{4}

3 reduction of

{Pb}^{2 +}

to

Pb

4 decomposition of

Pb

at the anode.

Explanation:

During charging of lead-acid battery ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on cathode.
At
Anode
${PbSO}_{4} (s) + 2 H_{2} O \to {PbO}_{2} (s) + {SO}_{4}^{2 -} (aq) + 4 H^{+} (aq) + 2 e^{-}$
At Cathode ${PbSO}_{4} (s) + 2 e^{-} \to Pb (s) + {SO}_{4}^{2 -} (aq)$
Thus, ${Pb}^{2 +}$ ions of ${PbSO}_{4}$ are reduced to $Pb$ on the cathode while ${PbSO}_{4}$ is oxidized to ${PbO}_{2}$ at anode.

AMU-2015

ELECTROCHEMISTRY

276257 The electrochemical cell stops working after sometime because :

1 electrode potential of both the electrodes becomes zero

2 electrode potential of both the electrodes becomes equal

3 one of the electrodes is eaten away

4 the cell reaction gets reversed

Explanation:

As we know that,
$Δ E_{cell} = E_{cathode} - E_{anode}$
when $E_{cathode} = E_{anode}$
$Δ E_{cell} = 0$
If $Δ E_{cell} = 0$
When electrode potential of both the electrodes becomes equal then the electrochemical cell stops working after sometime.

VITEEE- 2008

ELECTROCHEMISTRY

276258 The extent of charge of lead accumulator is determined by

1 amount of

{PbSO}_{4}

in the battery

2 amount of

{PbO}_{2}

in the battery

3 specific gravity of

H_{2} {SO}_{4}

of the battery

4 amount of

Pb

in the battery

Explanation:

The lead accumulator refers to a secondary cell as the electrical energy does not generate itself inside the cell, but it is stored prior to an external source.
The extent of charge of a lead accumulator is determined by the specific gravity of $H_{2} {SO}_{4}$ of the battery.
$\begin{matrix} Pb (s) + {PbO}_{2} (s) + 4 H^{+} (aq) + 2 {SO}_{4}^{2 -} (aq) ⟶ \\ 2 {PbSO}_{4} (l) + 2 H_{2} O \end{matrix}$
The fully charged lead accumulator is $1.30 g / ml$ and when it drops below $1.20 g / ml$ , the lead accumulator need recharging.

AP-EAMCET (Engg.) 2013

ELECTROCHEMISTRY

276262 Other things being equal, the EMF of a Daniel cell may be increased by

1 keeping low temperature

2 using large copper electrodes

3 using large zinc electrodes

4 decreasing concentration of

{Cu}^{2 +}

ions

Explanation:

For Daniel cell
$Zn + {Cu}^{2 +} \to {Zn}^{2 +} + Cu$
$E_{cell} = E_{cell}^{o} - \frac{RT}{nF} \ln \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}$
The decrease in temperature and increase in $[{Cu}^{2 +}]$ will increase the EMF.
Thus, low temperature will increases EMF of Daniel cell.

BCECE-2011

ELECTROCHEMISTRY

276264 In a Daniell cell constructed in the laboratory, the voltage observed was $0.9 V$ instead of $1.1 V$ of the standard cell. A possible explanation is-

1

[{Zn}^{2 +}] > [{Cu}^{2 +}]

2

[{Zn}^{2 +}] < [{Cu}^{2 +}]

3

Zn

electrode has twice the surface of

Cu

electrode

4 mol ratio of

{Zn}^{2 +} : {Cu}^{2 +}

is

2 : 1

Explanation:

$\begin{aligned} E_{cell} = E_{cell}^{\circ} - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ = 1.1 - \frac{0.0591}{2} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} \\ E_{cell} < 1.1 V if \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} > 1 \end{aligned}$
Thus, option (a) is correct.

BCECE-2013