The standard electrode potential $E^{\circ}$ and its temperature coefficient $\left(\frac{dE^{\circ}}{dT}\right)$ for a cell are 2 V and $-5 \times 10^{-4}$ VK$^{-1}$ at 300 K respectively. The cell reaction is: $$\ce{Zn(s) + Cu^{2+}(aq) Zn^{2+}(aq) + Cu(s)}$$ Standard reaction enthalpy ($\Delta_r H^{\circ}$) ______

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Redox Reactions and Electrochemistry

The standard electrode potential

E^{\circ}

and its temperature coefficient

(\frac{d E ^{\circ}}{d T})

for a cell are 2 V and

- 5 \times 1 0^{- 4}

^{- 1}

at 300 K respectively. The cell reaction is:

Zn (s) + Cu X^{2 +} (aq) Zn X^{2 +} (aq) + Cu (s)

Standard reaction enthalpy (

Δ_{r} H^{\circ}

) ______

Answer: (A)

Use the cell relations. The free-energy change is

Δ G^{\circ} = - n F E^{\circ} = - (2) (96500) (2) = - 386000 J = - 386 kJ

. The entropy change comes from the temperature coefficient,

Δ S^{\circ} = n F (\frac{d E ^{\circ}}{d T}) = (2) (96500) (- 5 \times 1 0^{- 4}) = - 96.5 J/K

. Then from

Δ H^{\circ} = Δ G^{\circ} + T Δ S^{\circ}

300 K

Δ H^{\circ} = - 386 + (300) (- 96.5 \times 1 0^{- 3}) = - 386 - 28.95 \approx - 412.8 kJ

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option