Question

Consider the following electrochemical cell \[ Zn | Zn^{2+}(aq) \, || \, Ag^{+}(aq) | Ag \] \[ [Zn^{2+}] = 1\,M \] The \(E_{cell\) was found to be \(1.6\,V\). The value of \(\log[Ag^+]\) is \[ E^\circ_{Zn^{2+}/Zn} = -0.76\,V , \qquad E^\circ_{Ag^+/Ag} = +0.80\,V \]

• A. \( \dfrac{1}{3} \)
• B. \( \dfrac{2}{3} \)
• C. \( \dfrac{3}{2} \)
• D. \( \dfrac{4}{3} \)

Answer 1

The Correct Option is B

Concept:
The cell potential under non-standard conditions is calculated using the Nernst equation: \[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n}\log Q \] where \(Q\) is the reaction quotient. Step 1: Write the overall cell reaction. \[ Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag \] Number of electrons transferred: \[ n = 2 \] Step 2: Calculate standard cell potential. \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] \[ E^\circ_{cell} = 0.80 - (-0.76) \] \[ E^\circ_{cell} = 1.56\,V \] Step 3: Write reaction quotient \(Q\). \[ Q = \frac{[Zn^{2+}]}{[Ag^+]^2} \] Since \( [Zn^{2+}] = 1 \): \[ Q = \frac{1}{[Ag^+]^2} \] Step 4: Apply Nernst equation. \[ E_{cell} = E^\circ_{cell} - \frac{0.059}{2}\log\left(\frac{1}{[Ag^+]^2}\right) \] \[ E_{cell} = E^\circ_{cell} + 0.059\log[Ag^+] \] Step 5: Substitute given values. \[ 1.6 = 1.56 + 0.059\log[Ag^+] \] \[ 0.04 = 0.059\log[Ag^+] \] \[ \log[Ag^+] \approx \frac{2}{3} \] \[ \boxed{\frac{2}{3}} \]