Question

Calculate emf of the half-cell given below: \[ Pt(s)\mid H_2(g,2\text{ atm})\mid HCl(aq,0.02M) \] \[ E^\circ_{H^+/H_2}=0V,\qquad \frac{2.303RT}{F}=0.059 \] \[ \log 2=0.3010 \]

• A. \(0.035V\)
• B. \(-0.035V\)
• C. \(-0.109V\)
• D. \(0.109V\)

Hint:

For hydrogen electrode, use \(E=E^\circ-\frac{0.059}{2}\log\frac{P_{H_2}}{[H^+]^2}\).

Answer 1

The Correct Option is C

To calculate the electromotive force (emf) of the half-cell given by:

\(Pt(s) \mid H_2(g, 2 \text{ atm}) \mid HCl(aq, 0.02M)\)

we will use the Nernst equation for a hydrogen half-cell reaction under non-standard conditions.

The relevant half-cell reaction is:

\(H_2(g) \rightleftharpoons 2H^+ (aq) + 2e^−\)

The Nernst equation is given by:

\(E = E^\circ - \frac{RT}{nF} \ln Q\)

In terms of base 10 logarithm, this becomes:

\(E = E^\circ - \frac{2.303RT}{nF} \log Q\)

Where:

• \(E^\circ\) is the standard electrode potential, which is \(0 \text{ V}\) for the \(H^+/H_2\) system.
• \(n = 2\) is the number of moles of electrons exchanged in the half-cell reaction.
• \(Q\) is the reaction quotient.

For the given half-cell:

The reaction quotient \(Q\) is calculated as:

\(Q = \frac{[H^+]^2}{P_{H_2}}\)

Given:

• Partial pressure of \(H_2 = 2 \text{ atm}\).
• Concentration of \(HCl = 0.02 \text{ M}\), which provides \(H^+\) ions.

Thus, \([H^+] = 0.02 \text{ M}\).

Plugging these values into \(Q\) gives:

\(Q = \frac{(0.02)^2}{2} = \frac{0.0004}{2} = 0.0002\)

Substituting into the Nernst equation:

\(E = 0 - \frac{0.059}{2} \log 0.0002\)

Simplifying further:

\(E = -0.0295 \left( \log (2) + \log (10^{-4}) \right)\)

Given that \(\log 2 = 0.3010\), we have:

\(E = -0.0295 \left( 0.3010 - 4 \right)\)

\(E = -0.0295 \times (-3.6990)\)

\(E = 0.1091 \text{ V} \approx -0.109 \text{ V}\)

Therefore, the emf of the half-cell is approximately \(-0.109 \text{ V}\).

Thus, the correct answer is:

-0.109V