\(\text{Given:}\) \[ E_{\text{Zn}}^\circ = -0.76 \, \text{V}, \, E_{\text{Cr}}^\circ = -0.74 \, \text{V} \] \(\text{To find: Emf of the cell } (E_{\text{cell}})\) \(\text{Formulae:}\) 1) \(E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ\) 2) \(E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0592 \, \text{V}}{n} \log_{10} \left[\frac{\text{Product}}{\text{Reactant}}\right]\) \(\text{Calculation:}\) \[ 3\text{Zn}_{(s)} \rightarrow 2\text{Cr}_{(0.1 \, \text{M})}^{3+} - 3\text{Zn}_{(0.08 \, \text{M})}^{2+} + \text{Cr}_{(s)} \, (\text{overall reaction}) \] Using formula (1): \[ E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ \] \[ E_{\text{cell}} = E_{\text{Cr}}^\circ - E_{\text{Zn}}^\circ \] \[ = -0.74 \, \text{V} - (-0.76 \, \text{V}) = 0.02 \, \text{V} \] Using formula (2): \[ \text{The cell potential is given by} \] \[ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.0592 \, \text{V}}{6} \log_{10} \left(\frac{0.08^3}{0.1^2}\right) \] \[ = 0.02 \, \text{V} + 0.0127 = 0.0327 \, \text{V} \] The emf of the cell is \(0.0327 \, \text{V}\).