Question

The reduction of O$_2$ to H$_2$O in acidic solution has a standard reduction potential of 1.23 V, during this process pH of solution is increased by 1 unit. Calculate the decrease in half cell potential (in mV) at 25°C. (Nearest integer) (Consider pressure of oxygen as 1 atm throughout the cell reaction)

Answer 1

59

Answer 2

The reduction half-cell reaction for O$_2$ to H$_2$O in acidic solution is:

$O_2(g) + 4H^+(aq) + 4e^- \longrightarrow 2H_2O(l)$

The Nernst equation at 25°C (298 K) is given by:

$E = E^o - \frac{0.0591}{n} \log Q$

For the given reaction:

1. Number of electrons (n): From the balanced equation, $n = 4$.

2. Reaction Quotient (Q):

   $Q = \frac{[H_2O]^2}{P_{O_2} [H^+]^4}$

   Since H$_2$O is a pure liquid, its activity is 1. The pressure of oxygen ($P_{O_2}$) is given as 1 atm.

   So, $Q = \frac{1}{1 \times [H^+]^4} = \frac{1}{[H^+]^4}$

Substitute these into the Nernst equation:

$E = E^o - \frac{0.0591}{4} \log \left(\frac{1}{[H^+]^4}\right)$

$E = E^o - \frac{0.0591}{4} \log ([H^+]^{-4})$

Using the logarithm property $\log(a^b) = b \log a$:

$E = E^o - \frac{0.0591}{4} (-4 \log [H^+])$

$E = E^o + 0.0591 \log [H^+]$

We know that $pH = -\log [H^+]$, which implies $\log [H^+] = -pH$.

Substitute this into the equation for E:

$E = E^o - 0.0591 \times pH$

Let the initial pH be $pH_1$ and the initial half-cell potential be $E_1$.

$E_1 = E^o - 0.0591 \times pH_1$

The pH of the solution is increased by 1 unit. So, the final pH, $pH_2$, is:

$pH_2 = pH_1 + 1$

The final half-cell potential, $E_2$, is:

$E_2 = E^o - 0.0591 \times pH_2$

The decrease in half-cell potential is $\Delta E = E_1 - E_2$.

$\Delta E = (E^o - 0.0591 \times pH_1) - (E^o - 0.0591 \times pH_2)$

$\Delta E = E^o - 0.0591 \times pH_1 - E^o + 0.0591 \times pH_2$

$\Delta E = 0.0591 \times pH_2 - 0.0591 \times pH_1$

$\Delta E = 0.0591 (pH_2 - pH_1)$

Substitute $pH_2 - pH_1 = 1$:

$\Delta E = 0.0591 \times 1$

$\Delta E = 0.0591 \text{ V}$

To express the decrease in millivolts (mV), multiply by 1000:

$\Delta E = 0.0591 \text{ V} \times 1000 \text{ mV/V} = 59.1 \text{ mV}$

Rounding to the nearest integer, the decrease in half-cell potential is 59 mV.