The standard `EMF` of quinhydrone is `0.699V`. The `EMF` of the quinhydrone electrode dipped in a solution with `pH=10` is
(a)`0.109V`
(b)`-0.109V`
(c)`1.289V`
(d)`-1.289V`

Options

A. `0.109V`
B. `-0.109V`
C. `1.289V`
D. `-1.289V`

Explanation

To find the EMF of the quinhydrone electrode in a solution with a pH of 10, we can use the Nernst equation. The steps are as follows:

### Step 1: Understand the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{1}{[H^+]} \right) \]
where:
- \( E \) is the electrode potential,
- \( E^\circ \) is the standard electrode potential (given as 0.699 V),
- \( n \) is the number of electrons transferred in the half-reaction (for quinhydrone, \( n = 1 \)),
- \( [H^+] \) is the concentration of hydrogen ions.

### Step 2: Calculate the Concentration of Hydrogen Ions
Given that the pH of the solution is 10, we can calculate the concentration of hydrogen ions:
\[ pH = -\log[H^+] \]
Thus,
\[ [H^+] = 10^{-pH} = 10^{-10} \, \text{M} \]

### Step 3: Substitute Values into the Nernst Equation
Now we can substitute the values into the Nernst equation:
\[ E = 0.699 - \frac{0.0591}{1} \log \left( \frac{1}{10^{-10}} \right) \]

### Step 4: Simplify the Logarithm
The logarithm can be simplified:
\[ \log \left( \frac{1}{10^{-10}} \right) = \log(10^{10}) = 10 \]

### Step 5: Substitute the Logarithm Value
Now substituting this back into the equation:
\[ E = 0.699 - 0.0591 \times 10 \]
\[ E = 0.699 - 0.591 \]
\[ E = 0.108 \, \text{V} \]

### Step 6: Round the Result
Rounding to three decimal places, we get:
\[ E \approx 0.109 \, \text{V} \]

### Conclusion
Thus, the EMF of the quinhydrone electrode dipped in a solution with pH 10 is:
**(a) 0.109 V**

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