Question
Chemistry Question on Nernst Equation
The electrode potential of the following half cell at 298K XX2+(0.001M)∥Y2+(0.01M)Y is _____×10−2V (Nearest integer)
Given: Ex2∗∣x0=−2.36V
EY0−1Y0=+0.36V
F2303RT=0.06V
Answer
The correct answer is 275.
X+Y2+→Y+X2+
E Cell0=0.36−(−2.36)=2.72V
ECell =2.72−20.06log0.010.001
=2.72+0.03=2.75V
=275×10−2V