For the following reaction-Pbs- Hg2SO4s - PbSO4s -2Hgl-Eocell-0-92 VKspPbSO4-2- 10-8- KspHg2SO4-1 - 10-6Hence- Ecell is-take -2 -1-4- log0-14 -0-85

We know, Q=[Pb^2+][Hg_2^2+] =√(K_sp(PbSO_4))√(K_sp(Hg_2SO_4)) =√(2 × 10^ 8)√(1 × 10^ 6) =√(2 × 10^ 2)=0.14 By using the relation, E_cell=E^o_cell 0.0592log 0.14 ...

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Byju's Answer

Standard XII

Chemistry

Gibb's Energy and Nernst Equation

For the follo...

Question

For the following reaction,
$P b (s) + H g_{2} S O_{4} (s) ⇌ P b S O_{4} (s) + 2 H g (l);$
$E_{c e l l}^{o} = 0.92 V$
$K_{s p} (P b S O_{4}) = 2 \times 10^{- 8}, K_{s p} (H g_{2} S O_{4}) = 1 \times 10^{- 6}$
Hence, $E_{c e l l}$ is:
(take $\sqrt{2} = 1.4, l o g 0.14 = - 0.85$ )

0.52 V

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0.68 V

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1.04 V

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0.95 V

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Solution

The correct option is D $0.95 V$
We know,
$Q = \frac{[P b^{2 +}]}{[H g_{2}^{2 +}]} = \frac{\sqrt{K_{s p} (P b S O_{4})}}{\sqrt{K_{s p} (H g_{2} S O_{4})}} = \frac{\sqrt{2 \times 10^{- 8}}}{\sqrt{1 \times 10^{- 6}}}$
$= \sqrt{2 \times 10^{- 2}} = 0.14$
By using the relation,
$E_{c e l l} = E_{c e l l}^{o} - \frac{0.059}{2} log 0.14$
Substituting the values, we get
$= 0.92 - \frac{0.059}{2} \times (- 0.85)$
$= 0.945 V \approx 0.95 V$

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Similar questions

P b (s) + H g_{2} S O_{4} ⇌ P b S O_{4} (s) + 2 H g (l)

design the cell if both electrolytes are present in their saturated solution state. Given

E_{P b / P b^{2 +}}^{o}

and

E_{H g / H g_{2}^{2 +}}^{o}

are 0.126 and -0.789 V respectively and K

_{s p}

of PbSO

_{4}

and Hg

_{2}

_{4}

are

2.43 \times 10^{- 8}

and

1.46 \times 10^{- 6}

respectively
The e.m.f. of given cell reaction to nearest integer

Q. The standard E.M.F of the cell written as

C d (H g) | C d S O_{4} . \frac{8}{3} H_{2} O (s) | | C d S O_{4} (s a t .) | H g_{2} S O_{4} (s) | H g

in which the cell reaction is

C d (H g) + H g_{2} S O_{4} (s) + \frac{8}{3} H_{2} O (l) ⇌ C d S O_{4} . \frac{8}{3} H_{2} O (s) + 2 H g (l)

1.0185 V

25^{o} C

. Calculate

Δ S^{o}

for the cell reaction if

{(\frac{\partial E^{o}}{\partial T})}_{P}

for the cell is

5.00 \times 10^{- 5} V K^{- 1}

Take,

F = 96485 C m o l^{- 1}

Q. If

K_{s p} (P b S O_{4}) = 1.8 \times 10^{- 8}

and

K_{a} (H S O_{4}^{⊝}) = 1.0 \times 10^{- 2}

for the following reaction, then the equilibrium constant for the reaction is

P b S O_{4} (s) + H^{\oplus} (a q) ⇌ H S O_{4}^{⊝} (a q) + P b^{2 +} (a q)

Q. At

298 K

, given that:

C u (s) | C u^{2 +} (1.0 M) | | A g^{+} (1.0 M) | A g (s)

E_{c e l l}^{o} = 0.46 V

Z n (s) | Z n^{2 +} (1.0 M) | | C u^{2 +} (1.0 M) | C u (s)

E_{c e l l}^{o} = 1.10 V

Then, the

E_{c e l l}

for the following reaction at

298 K

will be:

Z n (s) | Z n^{2 +} (0.1 M) | | A g^{+} (1. 0 M) | A g (s)

Q. If

K_{s p} (P b S O_{4}) = 1.8 \times 10^{- 8}

and

K_{a} (H S O_{4}^{⊝}) = 1.0 \times 10^{- 2}

, the equilibrium constant for the following reaction is

P b S O_{4} (s) + H^{\oplus} (a q) ⇌ H S O_{4}^{⊝} (a q) + P b^{2 +} (a q)

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For the following reaction, Pb(s)+Hg2SO4(s)⇌PbSO4(s)+2Hg(l); Eocell=0.92 V Ksp(PbSO4)=2×10−8, Ksp(Hg2SO4)=1×10−6 Hence, Ecell is: (take √2=1.4, log0.14=−0.85)

The correct option is D 0.95 VWe know, Q=[Pb2+][Hg2+2]=√Ksp(PbSO4)√Ksp(Hg2SO4)=√2×10−8√1×10−6 =√2×10−2=0.14 By using the relation, Ecell=Eocell−0.0592log0.14 Substituting the values, we get =0.92−0.0592×(−0.85) =0.945 V≈0.95 V

For the following reaction,
$P b (s) + H g_{2} S O_{4} (s) ⇌ P b S O_{4} (s) + 2 H g (l);$
$E_{c e l l}^{o} = 0.92 V$
$K_{s p} (P b S O_{4}) = 2 \times 10^{- 8}, K_{s p} (H g_{2} S O_{4}) = 1 \times 10^{- 6}$
Hence, $E_{c e l l}$ is:
(take $\sqrt{2} = 1.4, l o g 0.14 = - 0.85$ )