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प्रश्न
Calculate the emf and ΔG° for the cell reaction at 25°C:
\[\ce{\underset{(0.1 M)}{Zn_{(s)} | Zn^{2+}_{ (aq)}} || \underset{(0.01 M)}{Cd^{2+}_{ (aq)} | Cd_{(s)}}}\]
Given \[\ce{E^{\circ}_{Zn^{2+}/Zn}}\] = −0.763 and \[\ce{E^{\circ}_{Cd^{2+}/Cd}}\] = −0.403
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उत्तर
Given: The given electrochemical cell is:
\[\ce{\underset{(0.1 M)}{Zn_{(s)} | Zn^{2+}_{ (aq)}} | \underset{(0.01 M)}{Cd^{2+}_{ (aq)} | Cd_{(s)}}}\]
T = 25°C
= 25 + 273
= 298 K
\[\ce{E^{\circ}_{Zn^{2+}/Zn}}\] = −0.763 V,
\[\ce{E^{\circ}_{Cu^{2+}/Cu}}\] = −0.403 V
The cell reactions are:
\[\ce{Zn_{(s)} -> Zn{^{2+}_{(aq)}} + 2e-}\] (At anode)
\[\ce{Cd^{2+}_{ (aq)} + 2e- -> Cd_{(s)}}\] (At cathode)
\[\ce{Zn_{(s)} + Cd^{2+}_{ (aq)}-> Zn^{2+}_{ (aq)} + Cd_{(s)}}\] (Cell reaction)
n = 2
Nernst equation for the cell reaction is:
= \[\ce{E^{\circ}_{cell} + \frac{2.303 RT}{nF} log \frac{[Cd^{2+}]}{[Zn^{+2}]}}\]
= \[\ce{(-0.43 + 0.763) + \frac{0.0591}{2} log \frac{0.01}{0.1}}\]
= \[\ce{0.360 + \frac{0.0591}{2} log 10^{-1}}\]
= \[\ce{0.360 - \frac{0.0591}{2} log 10}\]
= \[\ce{0.360 - \frac{0.0591}{2} \times 1}\]
= 0.360 − 0.0296
= 0.330 V
\[\ce{\Delta G^{\circ} = -nFE^{\circ}_{cell}}\]
= −2 × 96500 × 0.360
= −0.72 × 96500
= −72 × 965
= −69480 J
ΔG° = −69.48 kJ
