Question

What is the \(E_{cell}\) (in V) of the following cell at 298 K?
\[ Zn(s) | Zn^{2+} (0.01 M) || Ni^{2+} (0.1 M) | Ni(s) \] Given \(E^\circ_{Zn^{2+}/Zn} = -0.76 V\), \(E^\circ_{Ni^{2+}/Ni} = -0.25 V\), \(\frac{2.303 RT}{F} = 0.06 V\)

• A. 0.51
• B. 0.48
• C. 0.57
• D. 0.54

Hint:

Use Nernst equation with reaction quotient to find cell potential.

Answer 1

The Correct Option is D

Step 1: Calculate standard cell potential
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-0.25) - (-0.76) = 0.51 V \] Step 2: Apply Nernst equation
\[ E_{cell} = E^\circ_{cell} - \frac{0.06}{n} \log Q \] where \(n=2\), reaction quotient: \[ Q = \frac{[Zn^{2+}]}{[Ni^{2+}]} = \frac{0.01}{0.1} = 0.1 \] Step 3: Calculate \(E_{cell}\)
\[ E_{cell} = 0.51 - \frac{0.06}{2} \log 0.1 = 0.51 - 0.03 \times (-1) = 0.54 V \] Step 4: Conclusion
Cell potential is 0.54 V.