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प्रश्न
Calculate the cell potential at 298 K:
\[\ce{Zn_{(s)} | Zn^{2+} (0.1 M) || Sn^{2+} (0.001 M) | Sn_{(s)}}\]
(Given: \[\ce{E{^{\circ}_{Zn^{2+}/Zn}}}\] = −0.76 V, \[\ce{E{^{\circ}_{Sn^{2+}/Sn}}}\] = −0.14 V, Gas constant, R = 8.314 J K−1 mol−1, Faraday constant, F = 96500 C mol−1)
संख्यात्मक
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उत्तर
Given: \[\ce{E{^{\circ}_{Zn^{2+}/Zn}}}\] = −0.76 V
\[\ce{E{^{\circ}_{Sn^{2+}/Sn}}}\] = −0.14 V
Gas constant, R = 8.314 J K−1 mol−1
Faraday constant, F = 96500 C mol−1
By using the Nernst equation
E° = `E_"cell"^circ - 0.0591/n log ((Zn^(2+))/(Sn^(2+)))`
= `0.62 - 0.0591/2 log (0.1/0.001)`
= 0.62 − 0.02955 × log(100)
= 0.62 − 0.02955 × 2
= 0.5609 V
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