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Question

Chemistry Question on Nernst Equation

The electrode potential of the following half cell at 298K298\, K XX2+(0.001M)Y2+(0.01M)YX \left| X ^{2+}(0.001 M ) \| Y ^{2+}(0.01 M )\right| Y is _____×102V\times 10^{-2} V (Nearest integer)
Given: Ex2x0=2.36VE _{ x ^{2 *} \mid x }^0=-2.36\, V
EY01Y0=+0.36VE _{ Y ^{0-1 Y }}^0=+0.36\, V
2303RTF=0.06V\frac{2303 RT }{ F }=0.06 \,V

Answer

The correct answer is 275.
X+Y2+→Y+X2+
E Cell0​=0.36−(−2.36)=2.72V
ECell ​=2.72−20.06​log0.010.001​
=2.72+0.03=2.75V
=275×10−2V