JEE Advanced levelSingle correct optionDifficulty 3 of 3ec166
A hydrogen-oxygen fuel cell at 298 K is first run with pure OX2 at 1bar and then with air, in which the partial pressure of OX2 is 0.21bar. Hydrogen is at 1bar in both cases and the product is liquid water. Taking F2.303RT=0.0591V and log2.1=0.322, the change in the open-circuit EMF on switching to air is
Aa decrease of about 10mV
Ba decrease of about 40mV
Can increase of about 10mV
Dno change, since the cathode reaction is the same
Answer: A
Cell reaction HX2+21OX2HX2O(l) with n=2 and Q=pHX2pOX21/21. Nernst: E=E∘−20.0591logQ=E∘+20.0591⋅21logpOX2. With pOX2=0.21, log0.21=0.322−1=−0.678, so ΔE=40.0591(−0.678)=−0.0100V, a drop of about 10mV. Writing the reaction per mole of OX2 (n=4, logpOX2) gives the same result.