Electrochemistry Question 9
Question 9 - 30 January - Shift 2
The electrode potential of the following half cell at 298 K.
$X|X^{2+}(0.001 M) | Y^{2+}(0.01 M)| Y$
is $\quad \times 10^{-2} V$ (Nearest integer).
Given : $E _{x^{2+} \mid x}^{0}=-2.36 V$
$E _{Y^{2+} \mid Y}^{0}=+0.36 V$
$\frac{2.303 RT}{F}=0.06 V$
Show Answer
Answer: (275)
Solution:
$X+Y^{2+} \to Y+X^{2+}$
$E _{\text{Cell }}^{0}=0.36-(-2.36)=2.72 V$
$ \begin{aligned} E _{\text{Cell }} & =2.72-\frac{0.06}{2} \log \frac{0.001}{0.01} \\ & =2.72+0.03=2.75 V \\ & =275 \times 10^{-2} V \end{aligned} $
