At 300 K- - H for the reaction-Zns - AgCl aq - ZnCl2aq - 2Ags is -218 kJ mol-1 while the EMF of the cell was 1-015 V- E- TP of the cell is

EMF of cell is given by nbsp; E= Δ H2F + T(∂ E∂ T)_P 1.015 V= 218 kJ mol^ 12× 96500 C mol^ 1 + 300 K(∂ E∂ T)_P 1.015 V=1.13 V+ 300 K(∂ E∂ T)_P 1.015 ...

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Standard XII

Chemistry

Arrhenius Equation

At 300 K, Δ H...

Question

At 300 K, $Δ H$ for the reaction:
$Z n (s) + A g C l (a q) \to Z n C l_{2} (a q) + 2 A g (s)$ is $- 218 k J m o l^{- 1}$ while the EMF of the cell was $1.015 V$ .
${(\frac{\partial E}{\partial T})}_{P}$ of the cell is

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

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

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1.10 \times 10^{- 1} V K^{- 1}

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7.62 \times 10^{- 2} V K^{- 1}

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Solution

The correct option is B $- 3.833 \times 10^{- 4} V K^{- 1}$
EMF of cell is given by
$E = - \frac{Δ H}{2 F} + T {(\frac{\partial E}{\partial T})}_{P}$
$1.015 V = - \frac{- 218 k J m o l^{- 1}}{2 \times 96500 C m o l^{- 1}} + 300 K {(\frac{\partial E}{\partial T})}_{P}$

$1.015 V = 1.13 V + 300 K {(\frac{\partial E}{\partial T})}_{P}$

$1.015 V - 1.13 V = 300 K {(\frac{\partial E}{\partial T})}_{P}$

$- 0.115 V = 300 K {(\frac{\partial E}{\partial T})}_{P}$
${(\frac{\partial E}{\partial T})}_{P} = - 3.833 \times 10^{- 4} V K^{- 1}$

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Similar questions

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:

Q. At

300 K

△ H

for the reaction :

Z n (s) + A g C l (s) ⟶ Z n C l_{2} (a q) + 2 A g (s)

- 218 k J / m o l

while the e.m.f. of the cell was

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

of the cell is:

Q. At 300K,

Δ

H for the reaction

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

, is

- 218

kJ/mol while the emf of the cell was

1.015 V

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

of the cell is :

For the cell reaction:

$Z n (s) + C u^{2 +} (0.1 M) \to Z n^{2 +} (0.01 M) + C u (s)$

If $E^{0}$ is the standard emf of the cell at 298 K, then:

Q. The standard emf of the cell,

C d_{(s)} | C d C l_{2 (a q)} (0.1 M) | | A g C l_{(a q)} | A g_{(s)}

In which the cell reaction is

C d_{(s)} + 2 A g C l_{(s)} \Rightarrow 2 A g_{(s)} + C d_{(a q)}^{2 +} + 2 C l_{(a q)}^{-}

0.6915 V

0^{o} C

and

0.6573

V

25^{o} C

The enthalpy change of the reaction at

25^{o} C

is:

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At 300 K, ΔH for the reaction: Zn(s)+AgCl(aq)→ZnCl2(aq)+2Ag(s) is −218 kJ mol−1 while the EMF of the cell was 1.015 V. (∂E∂T)P of the cell is

The correct option is B −3.833×10−4 V K−1 EMF of cell is given by E=−ΔH2F+T(∂E∂T)P 1.015 V=−−218 kJ mol−12×96500 C mol−1+300 K(∂E∂T)P 1.015 V=1.13 V+300 K(∂E∂T)P 1.015 V−1.13 V=300 K(∂E∂T)P −0.115 V=300 K(∂E∂T)P (∂E∂T)P=−3.833×10−4 V K−1

At 300 K, $Δ H$ for the reaction:
$Z n (s) + A g C l (a q) \to Z n C l_{2} (a q) + 2 A g (s)$ is $- 218 k J m o l^{- 1}$ while the EMF of the cell was $1.015 V$ .
${(\frac{\partial E}{\partial T})}_{P}$ of the cell is