Electrical energy produced by a reversible electrochemical cell is given by the free energy decrease - G of the reaction occurring in the cell- According to Gibbs- Helmholtz equation- decrease in free energy is given by - G- H - T - G - T - P where - H is the decrease in enthalpy of the cell reaction at constant pressure- EMF of the cell- E - H - nF - T - E - T - P- By measuring the emf of the cell and its temperature co-efficient- thermodynamic quantities like - H - - G and - S can be determined- Standard emf of the cell is related to equilibrium constant of the cell reaction as E 0-2-303 RT log k - nFZn - Zn -2- Fe -3- Fe -2- S - aq - aq - aq -A- 154-4 JK -B- -154-4 JK -C- 154-4 KJ K -D- 77-2 JK -

The correct option is A 154.4 JK^ [δ E/δ T ]_P=0.02/25VK^ 1 Δ S^∘ for the reaction =nF [δ E/δ T ]_P=2×96500×0.02/25 =154.4 JK^ 1 ...

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Standard XII

Chemistry

Introduction

Electrical en...

Question

Electrical energy produced by a reversible electrochemical cell is given by the free energy decrease $(- Δ G)$ of the reaction occurring in the cell. According to Gibbs- Helmholtz equation, decrease in free energy is given by $- Δ G = - Δ H - T {[\frac{δ (Δ G)}{δ T}]}_{P}$ where $- Δ H$ is the decrease in enthalpy of the cell reaction at constant pressure. EMF of the cell, $E = \frac{- Δ H}{n F} + T {[\frac{δ E}{δ T}]}_{P}$ . By measuring the emf of the cell and its temperature co-efficient, thermodynamic quantities like $Δ H, Δ G$ and $Δ S$ can be determined. Standard emf of the cell is related to equilibrium constant of the cell reaction as $E^{0} = \frac{2.303 R T l o g k}{n F}$ .

The standard emf of the cell, $\begin{matrix} Z n & Z n^{+ 2} & F e^{+ 3} F e^{+ 2} (S) & (a q) & (a q) (a q) \end{matrix}$ Pt at $25^{\circ}$ C is 1.53 V and at $50^{\circ}$ C is 1.55 V. The value of $Δ S^{\circ}$ for the overall reaction is

154.4 J K^{-}

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- 154.4 J K^{-}

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154.4 K J K^{-}

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77.2 J K^{-}

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Solution

The correct option is A $154.4 J K^{-}$
${[\frac{δ E}{δ T}]}_{P} = \frac{0.02}{25} V K^{- 1}$
$Δ S^{\circ}$ for the reaction $= n F {[\frac{δ E}{δ T}]}_{P} = 2 \times 96500 \times \frac{0.02}{25}$
$= 154.4 J K^{- 1}$

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Similar questions

Q. Electrical energy produced by a reversible electrochemical cell is given by the free energy decrease

(- Δ G)

of the reaction occurring in the cell. According to Gibbs- Helmholtz equation, decrease in free energy is given by

- Δ G = - Δ H - T {[\frac{δ (Δ G)}{δ T}]}_{P}

where

- Δ H

is the decrease in enthalpy of the cell reaction at constant pressure. EMF of the cell,

E = \frac{- Δ H}{n F} + T {[\frac{δ E}{δ T}]}_{P}

. By measuring the emf of the cell and its temperature co-efficient, thermodynamic quantities like

Δ H, Δ G

and

Δ S

can be determined. Standard emf of the cell is related to equilibrium constant of the cell reaction as

E^{0} = \frac{2.303 R T l o g k}{n F}

.

EMF of the cell,

\begin{matrix} A & A^{2 +} & B^{+} & B (S) & (a q) & (a q) & (s) 1 M & 0.1 M \end{matrix}

is found to be 1.475 Volt. The equilibrium constant of the cell reaction at

25^{\circ}

C is (approximately)

Q. Electrical energy produced by a reversible electrochemical cell is given by the free energy decrease

(- Δ G)

of the reaction occurring in the cell. According to Gibbs- Helmholtz equation, decrease in free energy is given by

- Δ G = - Δ H - T {[\frac{δ (Δ G)}{δ T}]}_{P}

where

- Δ H

is the decrease in enthalpy of the cell reaction at constant pressure. EMF of the cell,

E = \frac{- Δ H}{n F} + T {[\frac{δ E}{δ T}]}_{P}

. By measuring the emf of the cell and its temperature co-efficient, thermodynamic quantities like

Δ H, Δ G

and

Δ S

can be determined. Standard emf of the cell is related to equilibrium constant of the cell reaction as

E^{0} = \frac{2.303 R T l o g k}{n F}

.

From the following values of electrode potentials,
(i)

(f u m a r a t e)^{2 -} + 2 H^{+} + 2 e^{-} \to (s u c c i n a t e)^{- 2}, E_{1}^{0} = 0.03 V

and
(ii)

(p y r u v a t e)^{-} + 2 H^{+} + 2 e^{-} \to (l a c t a t e)^{-}, E_{2}^{0} = - 0.18 V . C a l c u l a t e Δ G^{\circ}

for the reaction,

(f u m a r a t e)^{2 -} + (l a c t a t e)^{-} \to (s u c c i n a t e)^{- 2} + (p y r u v a t e)^{-}

Q. The standard emf of the cell

F e | F e^{2 +} (a q) | | C d^{2 +} (a q) | C d

is 0.0372 V and temperature coefficient of the cell is

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

. Which is/are correct about the cell (at 300 K)?

Q. Consider an electrochemical cell:

A (s) | A^{n +} (a q, 2 M) | | B^{2 n +}

(aq, 1 M)|B(s). The value of

Δ H^{\circ}

for the cell reaction is twice that of

Δ G^{\circ}

at 300 K. If th emf of the cell is zero, the

Δ S^{\circ} (i n J K^{- 1} m o l^{- 1})

of the cell reaction per mole of B formed at 300 K is

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

(Given: ln(2) =0.7, R (universal gas constant) 8.3

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

. H, S and G are enthalpy, entropy and Gibbs energy, respectively.)

Q. If

Δ G = Δ H - T Δ S

and

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

then variation of EMF of a cell E, with temperature T, is given by:

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The correct option is A 154.4JK−[δEδT]P=0.0225VK−1 ΔS∘ for the reaction =nF[δEδT]P=2×96500×0.0225 =154.4JK−1

The correct option is A $154.4 J K^{-}$
${[\frac{δ E}{δ T}]}_{P} = \frac{0.02}{25} V K^{- 1}$
$Δ S^{\circ}$ for the reaction $= n F {[\frac{δ E}{δ T}]}_{P} = 2 \times 96500 \times \frac{0.02}{25}$
$= 154.4 J K^{- 1}$