In a cell, the following reactions take place \(Fe^{2+} → Fe^{3+} + e^-\) \(E°_{Fe^{3+}/Fe^{2+}} = 0.77\ V\) \(2I^- → I_2 + 2e^-\) \(E_{I_2/I^-} = 0.54\ V\) The standard electrode potential for the spontaneous reaction in the cell is \(x×10^{–2} V\) at \(208\ K\). The value of \(x\) is _______. (Nearest Integer)
To solve for the standard electrode potential of the spontaneous reaction in the cell, we need to determine the overall cell potential \(E°_{\text{cell}}\). The given half-reactions and their standard potentials are: \(Fe^{2+}→Fe^{3+}+e^-,\ E°_{Fe^{3+}/Fe^{2+}}=0.77\ V\) \(2I^-→I_2+2e^-,\ E_{I_2/I^-}=0.54\ V\) For the overall cell reaction, we need to balance the electron transfer. The first reaction gives off 1 electron, while the second reaction involves 2 electrons. Therefore, we will multiply the first reaction by 2 to equalize the electron transfer: \(2Fe^{2+}→2Fe^{3+}+2e^-\) Now sum both balanced reactions: \(2Fe^{2+}+2I^-→2Fe^{3+}+I_2\) The cell potential is calculated as: \(E°_{\text{cell}}=E°_{\text{cathode}}-E°_{\text{anode}}\) Assign the potential values: Cathode reaction (reduction): \(E°_{\text{cathode}}=0.54\ V\) Anode reaction (oxidation): \(E°_{\text{anode}}=0.77\ V\) Compute \(E°_{\text{cell}}\): \(E°_{\text{cell}}=0.54\ V-0.77\ V=-0.23\ V\) The problem statement indicates that \(E°_{\text{cell}}=x×10^{-2}\ V\) and asks for \(x\), where the value should be positive for consistency in expressing standard electrode potentials with spontaneous notation. Hence, \(-0.23\ V=23×10^{-2}\ V\). The absolute value of \(-23\) matches the provided range of 23 to 23. Thus, the nearest integer value of \(x\) is 23.
