MCQ
The oxidation potential of a hydrogen electrode at pH =10 and PH2 = 1 is:
- A0.059 V
- ✓+0.59 V
- C0.00 V
- D0.51 V
✓
Answer
Correct option: B.
+0.59 V
(B)+0.59 V
Explanation:
$\begin{array}{l}E_{OP}=E_{O P}^{\circ}-\frac{0.059}{1} \log \frac{\left[H^{+}\right]}{P_{H_2}} \\
\because\left[H^{+}\right]=10^{-10} ; P_{H_2}=1 atm \\
E_{OP}=0.59 V\end{array}$
Explanation:
$\begin{array}{l}E_{OP}=E_{O P}^{\circ}-\frac{0.059}{1} \log \frac{\left[H^{+}\right]}{P_{H_2}} \\
\because\left[H^{+}\right]=10^{-10} ; P_{H_2}=1 atm \\
E_{OP}=0.59 V\end{array}$
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