Question

The Gibb's energy change (in J) for the given reaction at [Cu²⁺] = [Sn²⁺] = 1 M and 298 K is ______.

\[\ce{Cu(s) + Sn^{2+}(aq) -> Cu^{2+}(aq) + Sn(s)}\]

`["E"_("Sn"^(2+)//"Sn")^0 = -0.16 "V", "E"_("Cu"^(2+)//"Cu")^0 = 0.34 "V", "Take F" = 96500  "C mol"^-1]`

Options

• 92000

• 94000

• 96500

• 98500

Answer 1

The Gibb's energy change (in J) for the given reaction at [Cu²⁺] = [Sn²⁺] = 1 M and 298 K is 96500.

\[\ce{Cu(s) + Sn^{2+}(aq) -> Cu^{2+}(aq) + Sn(s)}\]

`["E"_("Sn"^(2+)//"Sn")^0 = -0.16 "V", "E"_("Cu"^(2+)//"Cu")^0 = 0.34 "V", "Take F" = 96500  "C mol"^-1]`

Explanation:

Cell reaction:

Anode: \[\ce{Cu(s) -> Cu^{2+}(aq) + 2e^-}\]

Cathode: \[\ce{Sn^{2+}(aq) + 2e^- -> Sn(s)}\]

Overall reaction: \[\ce{Cu(s) + Sn^{2+}(aq) -> Cu^{2+}(aq) + Sn(g)}\]

`"E"_"cell"^0 = "E"_"cathode"^0 - "E"_"anode"^0`

= `"E"_("Sn"^(2+)//"Sn")^0 - "E"_("Cu"^(2+)//"Cu")^0`

= 0.16 − 0.34 V

= − 0.50 V

Gibbs energy change (ΔG) = `-"nF E"_"cell"^0`

= − 2 × 96500 C × −0.50 V

= + 96500 J