JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ec098
Consider the cell Pt∣HX2(1bar)∣HCl(0.01M)∣∣HCl(0.1M)∣HX2(1bar)∣Pt at 298 K. Take 2.303RT/F=0.0591V and F=96500Cmol−1. Which statements are correct?
AThe cell EMF is 0.0591V
BThe right hand electrode is the cathode
CΔG is about −5.70kJ per mole of HX+ transferred from the 0.1 M side to the 0.01 M side
DThe EMF becomes 0.1182V if the HX2 pressure on both sides is raised to 2 bar
Answer: A, B, C
Reduction 2HX++2eX−HX2 is favoured where [HX+] is higher, so the 0.1 M side (right) is the cathode (B). E=0.0591log0.010.1=0.0591V (A); using n=2 with [HX+]2 gives the same value. Net change per electron is HX+(0.1M)→HX+(0.01M), so ΔG=−1×96500×0.0591=−5703J≈−5.70kJ (C). Hydrogen pressure cancels when it is equal on both sides, so (D) is wrong.