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Saturday, 31 October

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JEE Main 8 January 2020, Shift 2, Chemistry Q48

Question 48 of 75 in this shift, Chemistry question 23 of 25, Section B.

For an electrochemical cell Sn(s)∣Sn2+(aq, 1M)∣∣Pb2+(aq, 1M)∣Pb(s)\mathrm{Sn(s)|Sn^{2+}(aq,\ 1M)||Pb^{2+}(aq,\ 1M)|Pb(s)} the ratio [Sn2+][Pb2+]\frac{[\mathrm{Sn^{2+}}]}{[\mathrm{Pb^{2+}}]} when this cell attains equilibrium is __________. (Given : ESn2+∣Sn0=−0.14 V, EPb2+∣Pb0=−0.13 V, 2.303RTF=0.06)\left(\text{Given : } \mathrm{E^0_{Sn^{2+}|Sn}} = -0.14\,\mathrm{V},\ \mathrm{E^0_{Pb^{2+}|Pb}} = -0.13\,\mathrm{V},\ \frac{2.303\mathrm{RT}}{\mathrm{F}} = 0.06\right)

Official answer

2.13 to 2.17

NTA final key (Jan 2020). Any value in the range scored.