Question

For the following question, enter the correct numerical value upto TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numeric value 5 will be written as 5.00 and 2.346 will be written as 2.35) Given the standard potential \( E^0_{\mathrm{Zn^{2+}/Zn}} = -0.76\,\mathrm{V} \), the EMF of the cell \[ \mathrm{Zn / Zn^{2+}_{(1M)} \parallel HCl \, (pH=2) \, | \, H_{2(atm)}} Pt \] is _____ V.

Hint:

For galvanic cells involving the standard hydrogen electrode, use the Nernst equation for the hydrogen half-cell: \( E = 0.0591 \log [\mathrm{H^+}] \) at 25°C and account for the pH.

Answer 1

Correct Answer: 0.6418

Step 1: The cell consists of a zinc electrode and a hydrogen electrode. The EMF of the cell is given by: \[ E_{\mathrm{cell}} = E^\circ_{\mathrm{Zn/Zn^{2+}}} - E_{\mathrm{H^+/H_2}} \]
Step 2: The potential of the hydrogen electrode is calculated using the Nernst equation: \[ E_{\mathrm{H^+/H_2}} = E^\circ_{\mathrm{H^+/H_2}} + \frac{0.0591}{1} \log [\mathrm{H^+}] \]
Step 3: Since \(E^\circ_{\mathrm{H^+/H_2}} = 0 \, \mathrm{V}\) and \(pH = 2\), the hydrogen ion concentration is: \[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-2} \, \mathrm{M} \]
Step 4: Substitute into the Nernst equation: \[ E_{\mathrm{H^+/H_2}} = 0 + 0.0591 \log(10^{-2}) = 0.0591 \times (-2) = -0.1182 \, \mathrm{V} \]
Step 5: Now calculate the EMF of the cell: \[ E_{\mathrm{cell}} = E^\circ_{\mathrm{Zn/Zn^{2+}}} - E_{\mathrm{H^+/H_2}} = (-0.76) - (-0.1182) = -0.76 + 0.1182 = -0.6418 \, \mathrm{V} \]
Step 6: By convention, the more positive electrode is considered the cathode. Since the hydrogen electrode has a more positive potential, it is the cathode. The final EMF is positive because the cell spontaneously generates current: \[ E_{\mathrm{cell}} = 0.6418 \, \mathrm{V} \]