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The electrochemical cells \(Zn| Zn^{2+} || Cu^{2+}| Cu\) and \(Fe | Fe^{2+} ||Cu^{2+} |Cu\) are connected in series. What will be the net emf of the cell at \(25^oC\)? Given: \(E^o\) of \(Zn^{2+}|Zn = -0.76 V\) \(E^o\) of \(Cu^{2+}|Cu = +0.34 V\) \(E^o\) of \(Fe^{2+}|Fe = -0.41 V\) |
+1.85 V -1.85 V +0.83 V 0.83 -V |
+1.85 V |
The correct answer is option 1. +1.85 V. To find the net electromotive force (emf) of the combined electrochemical cells connected in series, we need to consider the individual half-cell reactions and their standard reduction potentials (\(E^o\)). The net emf (\(E_{\text{cell}}\)) of the series combination is the sum of the emfs of the individual cells. Given the half-cell reactions and their standard reduction potentials: 1. For the cell \(Zn | Zn^{2+} || Cu^{2+} | Cu\): \(E^o_{\text{cell1}} = E^o_{\text{Cu}^{2+}/\text{Cu}} - E^o_{\text{Zn}^{2+}/\text{Zn}}\) \(E^o_{\text{cell2}} = E^o_{\text{Cu}^{2+}/\text{Cu}} - E^o_{\text{Fe}^{2+}/\text{Fe}}\) \(E^o_{\text{cell2}} = (+0.34 \, \text{V}) - (-0.41 \, \text{V}) = +0.75 \, \text{V}\) To find the net emf (\(E_{\text{cell}}\)) of the combined cells, we sum up the emfs of the individual cells: \(E_{\text{cell}} = E^o_{\text{cell1}} + E^o_{\text{cell2}}\) \(E_{\text{cell}} = +1.10 \, \text{V} + 0.75 \, \text{V} = +1.85 \, \text{V}\) So, the net emf of the cell at \(25^\circ C\) is \(+1.85 \, \text{V}\). Therefore, the correct answer is: 1. \(+1.85 \, \text{V}\) |