JEE MAIN - Chemistry (2019 - 9th April Morning Slot - No. 5)
The standard Gibbs energy for the given cell
reaction in kJ mol–1 at 298 K is :
Zn(s) + Cu2+ (aq) $$ \to $$ Zn2+ (aq) + Cu (s),
E° = 2 V at 298 K
(Faraday's constant, F = 96000 C mol–1)
Zn(s) + Cu2+ (aq) $$ \to $$ Zn2+ (aq) + Cu (s),
E° = 2 V at 298 K
(Faraday's constant, F = 96000 C mol–1)
384
–192
–384
192
Explanation
Here Zn is losing two electrons and Cu is gaining two electrons. So only two electrons are involved in the reaction.
$$ \therefore $$ n = 2
$$\Delta $$Go = - nFEo
= -2 $$ \times $$ 96000 $$ \times $$ 2
= -384 kJ/mol
$$ \therefore $$ n = 2
$$\Delta $$Go = - nFEo
= -2 $$ \times $$ 96000 $$ \times $$ 2
= -384 kJ/mol
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