Question

An electrochemical cell consists of pure Zn electrode (anode) and a hydrogen electrode (cathode) in a dilute Zn$^{+2}$ solution. The overall reaction is: \[ Zn(s) + 2H^+ \rightarrow H_2 + Zn^{+2} \] If the overall cell potential is +0.690 V, then the value of $\ln \left( \frac{[Zn^{+2}]}{[H^+]^2} \right)$ is ........... (rounded off to two decimal places). Given: Pressure of hydrogen gas = 1 atm; Temperature = 298 K; \[ \frac{RT}{F} = 0.0256 \, V \] Standard reduction potentials: \[ Zn^{+2} + 2e^- \rightarrow Zn (E^0 = -0.762 \, V) \] \[ 2H^+ + 2e^- \rightarrow H_2 (E^0 = 0 \, V) \]

Hint:

- If measured $E_{cell}<E^0_{cell}$, the reaction quotient $Q$ must be greater than 1 (more products than reactants). - Always divide $RT/F$ by $n$ (number of electrons) when applying the Nernst equation.

Answer 1

Correct Answer: 5.25

Step 1: Standard cell potential.
\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} = 0 - (-0.762) = 0.762 \, V \]
Step 2: Apply Nernst equation.
\[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \] Here, $n=2$, and \[ Q = \frac{[Zn^{+2}]}{[H^+]^2} \]
Step 3: Substitute experimental value.
\[ 0.690 = 0.762 - \frac{0.0256}{2} \ln Q \]
Step 4: Rearrangement.
\[ \frac{0.0256}{2} \ln Q = 0.762 - 0.690 = 0.072 \] \[ 0.0128 \ln Q = 0.072 \] \[ \ln Q = \frac{0.072}{0.0128} = 5.625 \]

Step 5: Interpretation.
Since $\ln Q>0$, this implies that $[Zn^{+2}]>[H^+]^2$. This means the solution is Zn$^{2+}$ rich compared to protons. Final Answer: \[ \boxed{5.63} \]