Consider the following cell reaction Cd s -Hg2SO4 s -9-5H2O l - CdSO4 -9-5H2O s -2Hg l The value of E0cell is 4-315 V at 250C - If - H0 - -825-2kJ mol-1 - the standard entropy change - S0 in J K-1 is - Nearest integer - Given -Faraday constant -96487 C mol -1 -

We know, Δ G= nFE^0 _cell Δ G=Δ H TΔ S nFE^0=Δ H TΔ S Δ S= nFE^0 Δ HT nbsp; = 2× 96487× 4.315+825.2× 10^3298= Δ S = 7482.81298 = 25.11 Δ S ≈25 J K^ ...

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Standard XII

Chemistry

Arrhenius Equation

Consider the ...

Question

Consider the following cell reaction

$C d_{(s)} + H g_{2} S O_{4 (s)} + \frac{9}{5} H_{2} O_{(l)} ⇌ C d S O_{4} . \frac{9}{5} H_{2} O_{(s)} + 2 H g_{(l)}$

The value of $E_{c e l l}^{0}$ is $4.315 V$ at $25^{0} C$ . If $Δ H^{0} = - 825.2$
kJ $m o l^{- 1}$ , the standard entropy change $Δ S^{0}$ in $J K^{- 1}$ is _________ . ( Nearest integer )

$[G i v e n : Faraday constant = 96487 C m o l^{- 1}]$

25.0

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25.00

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Solution

We know,
$Δ G = - n F E_{c e l l}^{0} Δ G = Δ H - T Δ S$
$- n F E^{0} = Δ H - T Δ S - Δ S = \frac{- n F E^{0} - Δ H}{T}$
$= \frac{- 2 \times 96487 \times 4.315 + 825.2 \times 10^{3}}{298} = - Δ S = - \frac{7482.81}{298} = - 25.11$

$Δ S \approx 25 J K^{- 1}$

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Similar questions

The standard entropy change (in

J K^{- 1} {m o l}^{- 1}

) for the given reaction is:

H_{2} (g) + {C l}_{2} (g) ⟶ 2 H C l (g)

25^{o} C

Express the value of the nearest integer of the standard entropy change in the form of

x + y

if the integer is

x y .

(Given

S^{o}

for

H_{2}, {C l}_{2}

and

H C l

are

0.13, 0.22

and

0.19 k J . K^{- 1} {m o l}^{- 1}

respectively.)

Q. The equilibrium constant for the reaction given below is

2.0 \times 10^{- 7}

at 300 K. Calculate the standard free energy change for the reaction.

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

Also, calculate the standard entropy change if,

Δ H^{\circ} = 28.40 k J m o l^{- 1}

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. The standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are 382.64 kJ mol

^{- 1}

and 145.6 J K

^{- 1}

mol

^{- 1},

respectively. Standard Gibb's energy change for the same reaction at 298 K is:

Standard enthalpy and standard entropy for the oxidation of $N H_{3}$ at $298$ K are $- 382.64$ $kJ {mol}^{- 1}$ and $- 145.6$ $J {mol}^{- 1} K^{- 1}$ respectively. Standard free energy change (in ${kJ mol}^{- 1}$ ) for the same reaction at $298$ K is:

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Consider the following cell reaction Cd(s)+Hg2SO4(s)+95H2O(l)⇌CdSO4 .95H2O(s)+2Hg(l) The value of E0cell is 4.315 V at 250C . If ΔH0=−825.2 kJ mol−1 , the standard entropy change ΔS0 in J K−1 is _________ . ( Nearest integer ) [Given :Faraday constant =96487 C mol−1]

We know, ΔG=−nFE0cellΔG=ΔH−TΔS −nFE0=ΔH−TΔS−ΔS=−nFE0−ΔHT =−2×96487×4.315+825.2×103298=−ΔS=−7482.81298=−25.11 ΔS≈25 J K−1

We know,
$Δ G = - n F E_{c e l l}^{0} Δ G = Δ H - T Δ S$
$- n F E^{0} = Δ H - T Δ S - Δ S = \frac{- n F E^{0} - Δ H}{T}$
$= \frac{- 2 \times 96487 \times 4.315 + 825.2 \times 10^{3}}{298} = - Δ S = - \frac{7482.81}{298} = - 25.11$

$Δ S \approx 25 J K^{- 1}$