Question

For the half cell

At $pH = 2$, the electrode potential is

• A. 1.36 V
• B. 1.30 V
• C. 1.42 V
• D. 1.20 V

Answer 1

Answer: (C)

In this electrode,
$[A]=[B]$ in quihydrone electrode
Hence, $Q=\left[ H ^{\oplus}\right]^2$
Hence, $Q=\left[ H ^{\oplus}\right]^2$

$ E =E^{\ominus}-\frac{0.0591}{2} \log \left[ H ^{\oplus}\right]^2 $
$=E^{\ominus}-0.0591 \log \left[ H ^{\oplus}\right] $
$=E^{\Theta_{+}} 0.0591 pH $
$=1.30+0.059 \times 2$
$= 1.42\, V $