JEE MAIN - Chemistry (2019 - 9th January Evening Slot - No. 11)
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction
Zn(s) + Cu2+ (aq) $$\rightleftharpoons$$ Zn2+(aq) + Cu(s)
at 300 K is approximately,
(R = 8 JK$$-$$1mol$$-$$1, F = 96000 C mol$$-$$1)
Zn(s) + Cu2+ (aq) $$\rightleftharpoons$$ Zn2+(aq) + Cu(s)
at 300 K is approximately,
(R = 8 JK$$-$$1mol$$-$$1, F = 96000 C mol$$-$$1)
e$$-$$80
e$$-$$160
e320
e160
Explanation
$$\Delta $$Go = $$-$$ RT lnk = $$-$$nFEocell
lnk = $${{n \times F \times {E^o}} \over {R \times T}} = {{2 \times 96000 \times 2} \over {8 \times 300}}$$
lnk = 160
k = e160
lnk = $${{n \times F \times {E^o}} \over {R \times T}} = {{2 \times 96000 \times 2} \over {8 \times 300}}$$
lnk = 160
k = e160
Comments (0)
