Question

If the electrode potential for a half-cell reaction at pH = 9, standard atmospheric pressure of oxygen 1 bar and 298K is given as x × 10⁻¹ volts. Then the value of x is _____. [Nearest integer]

$2H^+(aq) + \frac{1}{2}O_2(g) + 2e^- \rightarrow H_2O(l); E^o = 1.23V$

(R = 8.314 J mol⁻¹ K⁻¹)

Answer 1

7

Answer 2

The problem asks us to calculate the electrode potential for a given half-cell reaction at specific conditions and then find the value of 'x'.

The given half-cell reaction is: $2H^+(aq) + \frac{1}{2}O_2(g) + 2e^- \rightarrow H_2O(l)$

Given data: Standard electrode potential, $E^o = 1.23V$ Temperature, $T = 298K$ pH = 9 Standard atmospheric pressure of oxygen, $P_{O_2} = 1 \text{ bar}$ Number of electrons transferred, $n = 2$ (from the balanced half-reaction)

First, determine the hydrogen ion concentration $[H^+]$ from the pH: $\text{pH} = -\log_{10}[H^+]$ $9 = -\log_{10}[H^+]$ $[H^+] = 10^{-9} \text{ M}$

Next, write the expression for the reaction quotient ($Q$) for the given reaction: $Q = \frac{\text{Activity of products}}{\text{Activity of reactants}}$ For pure liquid water, its activity is 1. For gases, activity is approximated by partial pressure in bars. For aqueous species, activity is approximated by molar concentration. $Q = \frac{1}{[H^+]^2 \cdot (P_{O_2})^{1/2}}$

Substitute the values of $[H^+]$ and $P_{O_2}$ into the $Q$ expression: $Q = \frac{1}{(10^{-9})^2 \cdot (1)^{1/2}}$ $Q = \frac{1}{10^{-18} \cdot 1}$ $Q = 10^{18}$

Now, use the Nernst equation to calculate the electrode potential ($E$) at 298K: $E = E^o - \frac{0.0591}{n} \log_{10} Q$

Substitute the known values into the Nernst equation: $E = 1.23 - \frac{0.0591}{2} \log_{10} (10^{18})$ $E = 1.23 - \frac{0.0591}{2} \times 18$ $E = 1.23 - 0.0591 \times 9$ $E = 1.23 - 0.5319$ $E = 0.6981 \text{ V}$

The problem states that the electrode potential is $x \times 10^{-1}$ volts. So, $x \times 10^{-1} = 0.6981$ $x = \frac{0.6981}{10^{-1}}$ $x = 0.6981 \times 10$ $x = 6.981$

The question asks for the nearest integer value of x. The nearest integer to 6.981 is 7.

The final answer is $\boxed{7}$.

Explanation of the solution:

1. Determine $[H^+]$ from the given pH.
2. Write the expression for the reaction quotient $Q$ based on the given half-cell reaction.
3. Calculate the value of $Q$ using the calculated $[H^+]$ and given $P_{O_2}$.
4. Apply the Nernst equation ($E = E^o - \frac{0.0591}{n} \log_{10} Q$) to find the electrode potential $E$.
5. Equate the calculated $E$ to $x \times 10^{-1}$ and solve for $x$.
6. Round $x$ to the nearest integer.