The standard reduction potential at STP 25- C and 1 atm of the following half cells are found as below- Pb2-aq - 2e- - Pbs E0 - -0-121V PbO2s - 4H-aq - SO2-4aq - 2e- - PbSO4s - 2H2O E0 - 1-71 V Pb4-aq - 2e- - Pb2-aq E0 - 1-689 V For the cell PbSO4s - 2e- Pbs - SO2-4aq- The E0 is missing but it is found that in an aqueous solution of H2SO4 having a pH-1-7- when PbSO4s is put- then at saturated condition Solubility of PbSO4 - 5 - 10-5 Mol-L Take 2-303RT-F - 0-06 and log2 - 0-3 List-I contains questions and List-II contains their answers- - List-I List-II- What is the E0 in V for the half-cell - I PbSO4s - 2e- - Pbs - SO2-4aq - P -0-168- What is the E0cellin V of a lead storage battery- It has - II the cell reaction as - Pbs - PbO2s - 2H2SO4aq- 2PbSO4s - 2H2Ol Q 0-7- What is the E0 in V for the half cell - III PbO2s - 4H-aq - 4e- - Pbs - 2H2OI R 2-0- Ksp is the equilibrium constant for the reaction - IV PbO2s - 4H-aq- Pb4-aq - 2H2Ol S -0-068- What is the value of Eo in V- - T -0-31- U 2-02- - Which of the following options has the correct combination considering the reactions in List-I and their answers in List-II-

The correct option is B (I),(T) and (III),(Q) nbsp;For PbSO_4(s) + 2e^ → Pb_(s) + SO^2 _4(aq) E^0_1 Aqueous solution of H_2SO_4 having pH = 1.7 ⇒ [H^+ ...

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Standard XII

Chemistry

Nernst Equation

The standard ...

Question

The standard reduction potential at STP $(25^{\circ} C$ and 1 atm) of the following half cells are found as below.
$P b_{a q}^{2 +} + 2 e^{-} \to P b_{(s)} E^{0} = - 0.121 V P b O_{2 (s)} + 4 H_{(a q)}^{+} + S O_{4 (a q)}^{2 -} + 2 e^{-} \to P b S O_{4 (s)} + 2 H_{2} O E^{0} = 1.71 V P b_{(a q)}^{4 +} + 2 e^{-} \to P b_{(a q)}^{2 +} E^{0} = 1.689 V$
For the cell $P b S O_{4 (s)} + 2 e^{-} \to P b_{(s)} + S O_{4 (a q)}^{2 -}$ ,
The $E^{0}$ is missing but it is found that in an aqueous solution of $H_{2} S O_{4}$ having a pH=1.7, when $P b S O_{4 (s)}$ is put, then at saturated condition
Solubility of $P b S O_{4} = 5 \times 10^{- 5} M o l / L$
$(T a k e \frac{2.303 R T}{F} = 0.06 a n d l o g 2 = 0.3)$

List-I contains questions and List-II contains their answers.
$\begin{matrix} List-I & List-II What is the E^{0} (in V) for the half-cell (I) P b S O_{4 (s)} + 2 e^{-} \to P b_{(s)} + S O_{4 (a q)}^{2 -} ? & (P) -0.168 What is the E_{c e l l}^{0} (in V) of a lead storage battery? It has (II) the cell reaction as P b_{(s)} + P b O_{2 (s)} + 2 H_{2} S O_{4_{(a q)}} \to 2 P b S O_{4 (s)} + 2 H_{2} O_{(l)} & (Q) 0.7 What is the E^{0} (in V) for the half cell (III) P b O_{2 (s)} + 4 H_{(a q)}^{+} + 4 e^{-} \to P b_{(s)} + 2 H_{2} O_{(I)} & (R) 2.0 K_{s p} is the equilibrium constant for the reaction (IV) P b O_{2 (s)} + 4 H_{(a q)}^{+} ⇌ P b_{(a q)}^{4 +} + 2 H_{2} O_{(l)} & (S) -0.068 What is the value of E^{o} (in V) ? (T) -0.31 (U) 2.02 \end{matrix}$
Which of the following options has the correct combination considering the reactions in List-I and their answers in List-II?

(I), (S) and (III),(P)

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(I),(T) and (III),(Q)

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(I), (Q) and (III),(R)

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(I),(U) and (III),(R)

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Solution

The correct option is B (I),(T) and (III),(Q)

$F o r P b S O_{4 (s)} + 2 e^{-} \to P b_{(s)} + S O_{4 (a q)}^{2 -} E_{1}^{0}$
Aqueous solution of $H_{2} S O_{4} h a v i n g p H = 1.7 \Rightarrow [H^{+}] = 0.02 M$
So, $\Rightarrow [S O_{4}^{2 -}] = 0.01 M, a n d s o l u b i l i t y = 5 \times 10^{- 5} m o l / L$
$S o K_{s p} = 5 \times 10^{- 5} \times 0.01 = 5 \times 10^{- 7}$
So, at standard condition when $[S O_{4}^{2 -}] = 1 M t h e n [P b^{2 +}] = K_{s p} = 5 \times 10^{- 7} M$
So, from $P b_{(a q)}^{2 +} + 2 e^{-} \to P b_{(s)} E^{0} = - 0.121 V$
$E_{1}^{0} = - 0.121 - \frac{0.06}{2} l o g \frac{1}{K_{s p}} = - 0.121 - \frac{0.06}{2} l o g \frac{1}{5 \times 10^{- 7}} = - 0.31 V$
Now combining the cells
$P b O_{2 (s)} + 4 H_{(a q)}^{+} + S O_{4 (a q)}^{2 -} + 2 e^{-} \to P b S O_{4 (s)} + 2 H_{2} O E^{0} = 1.71 V A n d P b S O_{4 (s)} + 2 e^{-} \to P b_{(s)} + S O_{4 (a q)}^{2 -} E_{1}^{0} = - 0.31 V t o g e t P b_{(s)} + P b O_{2 (s)} + 2 H_{2} S O_{4 (a q)} \to 2 P b S O_{4 (s)} + 2 H_{2} O_{(I)} E_{C e l l}^{0} = 1.71 - (- 0.31) = 2.02 V$
Now adding $P b O_{2 (s)} + 4 H_{(a q)}^{+} + S O_{4 (a q)}^{2 -} + 2 e^{-} \to P b S O_{4 (s)} + 2 H_{2} O E^{0} = 1.71 V A n d P b S O_{4 (s)} + 2 e^{-} \to P b_{(s)} + S O_{4 (a q)}^{2 -} E_{1}^{0} = - 0.31 V t o g e t P b O_{2 (s)} + 4 H_{(a q)}^{+} + 4 e^{-} \to P b_{(s)} + 2 H_{2} O_{(I)}, S o E^{\circ} = \frac{1.71 \times 2 + (- 0.31) \times 2}{4} = 0.7 V$

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Similar questions

Q. The standard reduction potential at STP

(25^{\circ} C

and 1 atm) of the following half cells are found as below.

P b_{a q}^{2 +} + 2 e^{-} \to P b_{(s)} E^{0} = - 0.121 V P b O_{2 (s)} + 4 H_{(a q)}^{+} + S O_{4 (a q)}^{2 -} + 2 e^{-} \to P b S O_{4 (s)} + 2 H_{2} O E^{0} = 1.71 V P b_{(a q)}^{4 +} + 2 e^{-} \to P b_{(a q)}^{2 +} E^{0} = 1.689 V

For the cell

P b S O_{4 (s)} + 2 e^{-} \to P b_{(s)} + S O_{4 (a q)}^{2 -}

,
The

E^{0}

is missing but it is found that in an aqueous solution of

H_{2} S O_{4}

having a pH=1.7, when

P b S O_{4 (s)}

is put, then at saturated condition
Solubility of

P b S O_{4} = 5 \times 10^{- 5} M o l / L

(T a k e \frac{2.303 R T}{F} = 0.06 a n d l o g 2 = 0.3)

List-I contains questions and List-II contains their answers.

\begin{matrix} List-I & List-II What is the E^{0} (in V) for the half-cell (I) P b S O_{4 (s)} + 2 e^{-} \to P b_{(s)} + S O_{4 (a q)}^{2 -} ? & (P) -0.168 What is the E_{c e l l}^{0} (in V) of a lead storage battery? It has (II) the cell reaction as P b_{(s)} + P b O_{2 (s)} + 2 H_{2} S O_{4_{(a q)}} \to 2 P b S O_{4 (s)} + 2 H_{2} O_{(l)} & (Q) 0.7 What is the E^{0} (in V) for the half cell (III) P b O_{2 (s)} + 4 H_{(a q)}^{+} + 4 e^{-} \to P b_{(s)} + 2 H_{2} O_{(I)} & (R) 2.0 K_{s p} is the equilibrium constant for the reaction (IV) P b O_{2 (s)} + 4 H_{(a q)}^{+} ⇌ P b_{(a q)}^{4 +} + 2 H_{2} O_{(l)} & (S) -0.068 What is the value of E^{o} (in V) ? (T) -0.31 (U) 2.02 \end{matrix}

Which of the following options has the correct combination considering the reactions in List-I and their answers in List-II?

Q. The standard reduction potential at STP

(25^{\circ} C

and 1 atm) of the following half cells are found as below.

P b_{a q}^{2 +} + 2 e^{-} \to P b_{(s)} E^{0} = - 0.121 V P b O_{2 (s)} + 4 H_{(a q)}^{+} + S O_{4 (a q)}^{2 -} + 2 e^{-} \to P b S O_{4 (s)} + 2 H_{2} O E^{0} = 1.71 V P b_{(a q)}^{4 +} + 2 e^{-} \to P b_{(a q)}^{2 +} E^{0} = 1.689 V

For the cell

P b S O_{4 (s)} + 2 e^{-} \to P b_{(s)} + S O_{4 (a q)}^{2 -}

,
The

E^{0}

is missing but it is found that in an aqueous solution of

H_{2} S O_{4}

having a pH=1.7, when

P b S O_{4 (s)}

is put, then at saturated condition
Solubility of

P b S O_{4} = 5 \times 10^{- 5} M o l / L

(T a k e \frac{2.303 R T}{F} = 0.06 a n d l o g 2 = 0.3)

List-I contains questions and List-II contains their answers.

\begin{matrix} List-I & List-II What is the E^{0} (in V) for the half-cell (I) P b S O_{4 (s)} + 2 e^{-} \to P b_{(s)} + S O_{4 (a q)}^{2 -} ? & (P) -0.168 What is the E_{c e l l}^{0} (in V) of a lead storage battery? It has (II) the cell reaction as P b_{(s)} + P b O_{2 (s)} + 2 H_{2} S O_{4_{(a q)}} \to 2 P b S O_{4 (s)} + 2 H_{2} O_{(l)} & (Q) 0.7 What is the E^{0} (in V) for the half cell (III) P b O_{2 (s)} + 4 H_{(a q)}^{+} + 4 e^{-} \to P b_{(s)} + 2 H_{2} O_{(I)} & (R) 2.0 K_{s p} is the equilibrium constant for the reaction (IV) P b O_{2 (s)} + 4 H_{(a q)}^{+} ⇌ P b_{(a q)}^{4 +} + 2 H_{2} O_{(l)} & (S) -0.068 What is the value of E^{o} (in V) ? (T) -0.31 (U) 2.02 \end{matrix}

Which of the following options has the correct combination considering the reactions in List-I and their answers in List-II?

Q. Match the column I with column II and mark the appropriate choice.

	Column I		Column II
(A)	$P b_{(s)} + S O_{4 (a q)}^{- 2} \to P b S O_{4 (s)} + 2 e^{-}$	(i)	Rusting of iron
(B)	$2 S O_{4 (a q)}^{2 -} \to S_{2} O_{8 (a q)}^{2 -} + 2 e^{-}$	(ii)	Reaction at anode in lead storage battery
(C)	$2 H_{2 (g)} + 4 O H_{(a q)}^{-} \to 4 H_{2} O_{(l)} + 4 e^{-}$	(iii)	Electrolysis of concentrated $H_{2} S O_{4}$
(D)	$2 F e_{(s)} + O_{2 (g)} + 4 H_{(a q)}^{+} \to 2 F e_{(a q)}^{2 +} + 2 H_{2} O_{(l)}$	(iv)	Reaction at anode in fuel cell

Q. Which of the following metals will couple to provide a higher cell potential for a galvanic cell?

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

N i^{2 +} (a q) + 2 e^{-} \to N i (s); E^{0} = - 0.25 V

P b^{2 +} (a q) + 2 e^{-} \to P b (s); E^{0} = - 0.13 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

Q. Which of the following metals, when coupled, will leads to a non-spontaneous cell reaction?
Given:

A u^{+} (a q) + e^{-} \to A u (s); E^{0} = 1.69 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

P b^{2 +} (a q) + 2 e^{-} \to P b (s); E^{0} = - 0.13 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

C a^{2 +} (a q) + 2 e^{-} \to C a (s); E^{0} = - 2.87 V

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