Question

Calculate the EMF of the cell:
Zn(s) | Zn\(^{2+}\) (0.1 M) || Cu\(^{2+}\) (1.0 M) | Cu(s)
Given: \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,V \), \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\,V \)

Hint:

Always place the less reactive metal on the left in the cell notation. Use the Nernst equation for non-standard conditions.

Answer 1

Step 1: Standard EMF of the cell: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = 1.10\,V \] Step 2: Apply Nernst equation: \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \] Here, \( n = 2 \) \[ E_{cell} = 1.10 - \frac{0.0591}{2} \log \left( \frac{0.1}{1.0} \right) = 1.10 - 0.02955 \cdot (-1) = 1.10 + 0.02955 = 1.1296\,V \] Final Answer: \[ \boxed{E_{cell} = 1.13\,V} \]