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Question
Calculate the value of ΔG° for the following cell at 25°C.
\[\ce{Zn_{(s)}/Zn^2+(1M) || Sn^2+(1M)/Sn_{(s)}}\]
Given: \[\ce{E^{\circ}_{({Zn^{2+}/{Zn}})}}\] = −0.76 V, \[\ce{E^{\circ}_{({Sn^{2+}/{Sn}})}}\] = −0.14 V
1 faraday = 96,500 coulombs.
Numerical
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Solution
Given: \[\ce{E^{\circ}_{({Zn^{2+}/{Zn}})}}\] = −0.76 V
\[\ce{E^{\circ}_{({Sn^{2+}/{Sn}})}}\] = 0.14 V
Formula: \[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]
= −0.14 − (−0.76)
= −0.14 + 0.76
= 0.62 V
ΔG° = −nFE°cell
= −2 × 96500 × 0.62
= −119,660 J/mol
ΔG° = `(−119,660)/(1000)`
= −119.66 kJ/mol
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