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 \(Fe^{2+}|Fe = -0.41 V\)
Answer & explanation
Correct answer: option 1
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 \(Zn | Zn^{2+} || Cu^{2+} | Cu\):
\(E^o_{\text{cell1}} = E^o_{\text{Cu}^{2+}/\text{Cu}} - E^o_{\text{Zn}^{2+}/\text{Zn}}\)
\(E^o_{\text{cell1}} = (+0.34 \, \text{V}) - (-0.76 \, \text{V}) = +1.10 \, \text{V}\)
2. For the cell \(Fe | Fe^{2+} || Cu^{2+} | Cu\):
\(E^o_{\text{cell2}} = E^o_{\text{Cu}^{2+}/\text{Cu}} - E^o_{\text{Fe}^{2+}/\text{Fe}}\)
\(E^o_{\text{cell2}} = (+0.34 \, \text{V}) - (-0.41 \, \text{V}) = +0.75 \, \text{V}\)
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. \(+1.85 \, \text{V}\)