JEE PYQ: Electrochemistry Question 34
Question 34 - 2019 (10 Jan Shift 2)
In the cell Pt(s) | H$_2$(g, 1bar) / HCl(aq)||AgCl(s)/Ag(s)|Pt(s), the cell potential is 0.92 V when a $10^{-6}$ molal HCl solution is used. The standard electrode potential of (AgCl/Ag, Cl$^-$) electrode is:
$\left{\text{Given}: \frac{2.303RT}{F} = 0.06 \text{ V at 298 K}\right}$
(1) 0.94 V
(2) 0.76 V
(3) 0.40 V
(4) 0.20 V
Show Answer
Answer: (4) 0.20 V
Solution
$0.92 = E^0 - 0.06\log 10^{-12}$. $E^0 = 0.92 - 0.72 = 0.20$ V.
BETA
