\(\text{Given:}\) \[ [Zn^{2+}] = 0.2 \, \text{M}, \, [H^+] = 1.6 \, \text{M}, \, P_{H_2} = 1.8 \, \text{atm} \] \(\text{To find: Emf of the cell } (E_{\text{cell}})\) \(\text{Formulae:}\) 1) \(E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ\) 2) \(E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0592 \, \text{V}}{n} \log_{10} \left[\frac{\text{Product}}{\text{Reactant}}\right]\) \(\text{Calculation:}\) \[ \text{Zn}_{(s)} \rightarrow Zn_{(0.2 \, \text{M})}^{2+} + 2e^{-} \, (\text{oxidation at anode}) \] \[ 2H_{(1.6 \, \text{M})}^{+} + 2e^{-} \rightarrow H_{2(1.8 \, \text{atm})} \, (\text{reduction at cathode}) \] \(\text{Overall reaction:}\) \[ \text{Zn}_{(s)} + 2H_{(1.6 \, \text{M})}^{+} \rightarrow Zn_{(0.2 \, \text{M})}^{2+} + H_{2(1.8 \, \text{atm})} \] \[ E_{H_2}^\circ = 0.0 \, \text{V} \, \text{and} \, E_{Zn}^\circ = -0.763 \, \text{V} \] Using formula (1): \[ E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ \] \[ E_{\text{cell}} = E_{H_2}^\circ - E_{Zn}^\circ = 0.0 \, \text{V} - (-0.763 \, \text{V}) = 0.763 \, \text{V} \] Using formula (2): \[ \text{The cell potential is given by} \] \[ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0592 \, \text{V}}{2} \log_{10} \left[\frac{(0.2)(1.8)}{(1.6)^2}\right] \] \[ = 0.763 - 0.0252 = 0.7882 \, \text{V} \] The emf of the cell is \(0.7882 \, \text{V}\).