The cell potential E cell of a reaction is related as - G - nF E cell- where - G represents maximum useful electrical work- n - no- of moles of electrons exchanged during the reaction- For reversible cell reaction d - G - V dp - S d TAt const- P- d - G - S dTAt const- P- - G - H - T - S- G - H - T d - G - T P - dE cell - dT p is known as temperature coefficient of the emf of the cell-At 300 K - - H for the reaction Zn -2 AgCl - ZnCl 2-2 Agis -218 kJ - mole while emf of the cell was 1-015 V- left fracdE dT right pl of the cell is A- -4-2 - 10-4 V K -1B- -3-81 - 10-4 V K -1C- 0-11 V K -1D- 7-62 - 10-4 V K -1

The correct option is B 3.81×10^ 4VK^ 1Δ H=Δ G+TΔ S = nFE+nF ( dE/dT )T 218× 10^3= 2× 96500 × 1.015+2×96500× 300 ( dE/dT )_p dE/dT= 3.81×10^ 4VK^ 1 ...

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Byju's Answer

Standard XII

Chemistry

Properties of Water

The cell pote...

Question

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

- 4.2 \times 10^{- 4} V K^{- 1}

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

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0.11 V K^{- 1}

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7.62 \times 10^{- 4} V K^{- 1}

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Solution

The correct option is B $- 3.81 \times 10^{- 4} V K^{- 1}$
$Δ H = Δ G + T Δ S$
$= - n F E + n F (\frac{d E}{d T}) T$
$- 218 \times 10^{3} = - 2 \times 96500 \times 1.015 + 2 \times 96500 \times 300 {(\frac{d E}{d T})}_{p}$
$\frac{d E}{d T} = - 3.81 \times 10^{- 4} V K^{- 1}$

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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.

The

Δ S

for the cell reaction

Z n + 2 A g C l \to Z n C l_{2} + 2 A g

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.

The temperature coefficient of the emf the 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}

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}

Calculate

Δ S

for the given cell reaction in the previous question.

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The correct option is B −3.81×10−4VK−1ΔH=ΔG+TΔS =−nFE+nF(dEdT)T −218×103=−2×96500×1.015+2×96500×300(dEdT)p dEdT=−3.81×10−4VK−1

The correct option is B $- 3.81 \times 10^{- 4} V K^{- 1}$
$Δ H = Δ G + T Δ S$
$= - n F E + n F (\frac{d E}{d T}) T$
$- 218 \times 10^{3} = - 2 \times 96500 \times 1.015 + 2 \times 96500 \times 300 {(\frac{d E}{d T})}_{p}$
$\frac{d E}{d T} = - 3.81 \times 10^{- 4} V K^{- 1}$