Question

The emf of a standard cadmium cell is 1.02 V at 300 K. The temperature coefficient of the cell
is –5.0×10–5 V K–1
. The value of H° for the cell is _________ kJ mol–

Answer 1

The value of $\Delta H^o$ is $-199.75\ kJ\ mol^{-1}$ .

Given :

Emf , $E^o=1.02\ V$ .

Temperature , T = 300 K .

Temperature coefficient of the cell  is , $\dfrac{\partial E}{\partial T}=-5\times 10^{-5}\ VK^{-1}$

We need to find $\Delta H^o$ for the cell .

We know , emf is given as :

$E^o=-\dfrac{\Delta H^o}{nF}+T(\dfrac{\partial E}{\partial T})_P$

Putting all the value in above equation , we get :

$\Delta H^o=-199.75\ kJ\ mol^{-1}$

Hence , this is the required solution .

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Answer 2

The value of ΔH° for the cell is -199.75 kJ mol⁻¹

Given:

E° = 1.02 V

T = 300 K

∂E/∂T = –5.0 × 10⁻⁵ VK⁻¹

To find:

ΔH° = ?

Explanation:

The emf of the standard cell is given by the formula:

E° = - ΔH°/nF + T(∂E/∂T)

On substituting the values, we get,

1.02 = - ΔH°/(2 × 96500) + 300(–5.0 × 10⁻⁵)

ΔH°/(2 × 96500) = -1.02 - 300(–5.0 × 10⁻⁵)

ΔH° = (2 × 96500) × (-1.02 - 300(–5.0 × 10⁻⁵))

ΔH° = 193000 (-1.02 - 0.015)

ΔH° = 193000 (-1.035)

∴ ΔH° = - 199755 Jmol⁻¹ = - 199.755 kJ mol⁻¹