Question

Calculate the values of E_cell and ΔG for the following cell reaction at 25°C:

\[\ce{Zn_{(s)}/Zn^{2+}_{(0.0004 M)} || Cd^{2+}_{(0.2 M)}/Cd_{(s)}}\]

(Given, \[\ce{E^°_{Zn^{2+}/Zn}}\] = −0.763 V and \[\ce{E^°_{Cd^{2+}/Cd}}\] = −0.403 V, 1 Faraday = 96,500 coulombs, R = 8.314 JK⁻¹ mol⁻¹ )

Answer 1

Given: \[\ce{Zn_{(s)}/Zn^{2+}_{(0.0004 M)} || Cd^{2+}_{(0.2 M)}/Cd_{(s)}}\]

\[\ce{E^°_{Zn^{2+}/Zn}}\] = −0.763 V

\[\ce{E^°_{Cd^{2+}/Cd}}\] = −0.403 V

Cell reaction: \[\ce{Zn_{(s)} + Cd^{2+}_{ (aq)}-> Zn^{2+}_{ (aq)} + Cd_{(s)}}\]

\[\ce{E^° = E^°_{cathode} - E^°_{anode}}\]

= −0.403 V − (−0.763 V)

= 0.36 V

Nernst equation for the cell reaction is:

\[\ce{E_{cell} = E^°_{cell} - \frac{0.059 }{2} log \frac{[Zn^{+2}]}{[Cd^{2+}]}}\]

= `0.36 - 0.059/2 log  ([0.0004])/([0.2])` 

= `0.36 - 0.059/2 log  ([2])/([10^(3)])` 

= `0.36 - 0.059/2 [log (2) - log (10^(3))]` 

= 0.36 − 0.0295 [0.3010 − 3]

= 0.36 − 0.0295 (−2.6990)

= 0.36 − (−0.079)

= 0.36 + 0.079

= 0.36 + 0.08

= 0.44 V

E_cell = + 0.44 V

ΔG = – nFE_cell

= –2 × 96500 × 0.44

= – 84920 J

= – 84.920 kJ