JEE MAIN - Chemistry (2019 - 12th January Morning Slot - No. 7)
The standard electrode potential $${E^o }$$ and its temperature coefficient $$\left( {{{d{E^o }} \over {dT}}} \right)$$ for a cell are 2V and $$-$$ 5 $$ \times $$ 10$$-$$4 VK$$-$$1 at 300 K respectively.
The cell reaction is
Zn(s) + Cu2+ (aq) $$\buildrel \, \over \longrightarrow $$ Zn2+ (aq) + Cu(s)
The standard reaction enthalpy ($$\Delta $$rH$${^o }$$) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]
The cell reaction is
Zn(s) + Cu2+ (aq) $$\buildrel \, \over \longrightarrow $$ Zn2+ (aq) + Cu(s)
The standard reaction enthalpy ($$\Delta $$rH$${^o }$$) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]
$$-$$ 412.8
$$-$$ 384.0
192.0
206.4
Explanation
$$\Delta $$G = -nFEcell = -2$$ \times $$96500$$ \times $$2 = -386 kJ
$$\Delta $$S = nF$$\left( {{{d{E^o }} \over {dT}}} \right)$$ = 2$$ \times $$96500$$ \times $$ ($$-$$ 5 $$ \times $$ 10$$-$$4) = -96.5 kJ
At 298 K
T$$\Delta $$S = 298 $$ \times $$ (–96.5 J) = – 28.8 kJ
at constant T (=248 K) and pressure
$$\Delta $$G = $$\Delta $$H – T$$\Delta $$S
$$ \Rightarrow $$ $$\Delta $$H = $$\Delta $$G + T$$\Delta $$S
= -386 - 28.8 = -412.8 kJ
$$\Delta $$S = nF$$\left( {{{d{E^o }} \over {dT}}} \right)$$ = 2$$ \times $$96500$$ \times $$ ($$-$$ 5 $$ \times $$ 10$$-$$4) = -96.5 kJ
At 298 K
T$$\Delta $$S = 298 $$ \times $$ (–96.5 J) = – 28.8 kJ
at constant T (=248 K) and pressure
$$\Delta $$G = $$\Delta $$H – T$$\Delta $$S
$$ \Rightarrow $$ $$\Delta $$H = $$\Delta $$G + T$$\Delta $$S
= -386 - 28.8 = -412.8 kJ
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