D - G - S reaction - dT - G - H - T - S - G - H - T d - G - dT PAt 300 K - - H for the reaction-is -218 kJ - mol while the e-m-f of the cell is 1-015 V- dE - dT P of the cell is-A- -4-2 - 10-4 VK -1B- -3-81 - 10-4 VK -1C- 7-62 - 10-4 VK -1D- 0-11 VK -1

The correct option is B 3.81× 10^ 4 VK^ 1d(Δ G) = Δ S dT Δ G = Δ H TΔ S Δ G = Δ H + T d(Δ G)/dT Δ G = Δ H + T d( nFE)/dT Δ G = Δ H nFT dE/dT dE/dT = Δ G ...

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Byju's Answer

Standard XII

Chemistry

Gibb's Free Energy

d Δ G =-Δ S r...

Question

$d (Δ G) = - (Δ S_{reaction}) . d T$ $Δ G = Δ H - T Δ S$ $Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}$
At $300 K, Δ H$ for the reaction,
$Z n_{(S)} + 2 A g C l_{(S)} \to Z n C l_{2_{(a q)}} + 2 A g_{(S)}$
is $- 218 kJ/mol$ while the e.m.f of the cell is $1.015 V$ . ${(\frac{d E}{d T})}_{P}$ of the cell is:

- 4.2 \times 10^{- 4} {VK}^{- 1}

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- 3.81 \times 10^{- 4} {VK}^{- 1}

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0.11 {VK}^{- 1}

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7.62 \times 10^{- 4} {VK}^{- 1}

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Solution

The correct option is B $- 3.81 \times 10^{- 4} {VK}^{- 1}$
$d (Δ G) = - Δ S d T$

$Δ G = Δ H - T Δ S$

$Δ G = Δ H + T \frac{d (Δ G)}{d T}$

$Δ G = Δ H + T \frac{d (- n F E)}{d T}$

$Δ G = Δ H - n F T \frac{d E}{d T}$

$\frac{d E}{d T} = \frac{Δ G - Δ H}{- n F T}$

$\frac{d E}{d T} = \frac{- n F E - Δ H}{- n F T}$

Also, $n = 2$

$\frac{d E}{d T} = \frac{- 2 \times 96500 \times 1.015 + 218000}{- 2 \times 96500 \times 300}$

$= - 3.81 \times 10^{- 4} V K^{- 1}$

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Similar questions

Q. The cell potential

(E_{c e l l})

of a reaction is related as

Δ G = n F E_{c e l l}

, where

Δ G

represents maximum useful electrical work.
n = no. of moles of electrons exchanged during the reaction.
For reversible cell reaction

d (Δ G) = (Δ V) d p - (Δ S) d T

At const. P,

d (Δ G) = - (Δ S) d T

At const. P,

Δ G = Δ H - T Δ S

. (1)

∴ Δ G = Δ H + T {(\frac{d (Δ G)}{Δ T})}_{P}

. (2)

{(\frac{d E_{c e l l}}{d T})}_{p}

is known as temperature coefficient of the emf of the cell.

At 300K,

Δ H

for the reaction

Z n + 2 A g C l \to Z n C l_{2} + 2 A g i s - 218

kJ/mole while emf of the cell was

1.015 V . {(\frac{d E}{d T})}_{p}

of the cell is

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

The temperature coefficient of the e.m.f of a cell,

{(\frac{d E}{d T})}_{P}

is given by:

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

300 K, Δ H

for the reaction,

Z n_{(S)} + 2 A g C l_{(S)} \to Z n C l_{2_{(a q)}} + 2 A g_{(S)}

- 218 kJ/mol

while the e.m.f of the cell is

1.015 V

{(\frac{d E}{d T})}_{P}

of the cell is:

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

The temperature coefficient of the e.m.f of a cell,

{(\frac{d E}{d T})}_{P}

is given by:

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

Calculate

Δ S

for the given cell reaction in the previous question.

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d(ΔG)=−(ΔSreaction).dT ΔG=ΔH−TΔS ΔG=ΔH+T(d(ΔG)dT)P At 300 K,ΔH for the reaction, Zn(S)+2AgCl(S)→ZnCl2(aq)+2Ag(S) is −218 kJ/mol while the e.m.f of the cell is 1.015 V. (dEdT)P of the cell is: