9) Which of the following(s) is/are correct for the following cell? Cd^+2(0.25M)vert Cd(s)Vert Cu^+2(0.45M)vert Cu(s) (E^circ Cu^+2/Cu=0.337V;E^circ Cu^+2/Cd=-0.403V) I- The cell type is an electrolytic cell II- Cell Voltage is -0.421V III- The Free energy of the reaction is -144.25kJ A) I and II B) I and III C II and III D)Only I Only III

To determine which statements are correct for the given cell, we need to analyze the cell notation and the provided standard reduction potentials.<br /><br />The given cell notation is:<br />$Cd^{+2}(0.25M)\vert Cd(s)\Vert Cu^{+2}(0.45M)\vert Cu(s)$<br /><br />I- The cell type is an electrolytic cell.<br />This statement is incorrect. The given cell is a galvanic (or voltaic) cell, not an electrolytic cell. In a galvanic cell, the spontaneous redox reaction generates an electric current.<br /><br />II- Cell Voltage is $-0.421V$.<br />To calculate the cell voltage, we can use the Nernst equation:<br />$E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log Q$<br /><br />Where:<br />$E^{\circ}_{cell}$ is the standard cell potential, which can be calculated using the standard reduction potentials:<br />$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$<br />$E^{\circ}_{cell} = 0.337V - (-0.403V) = 0.740V$<br /><br />$n$ is the number of moles of electrons transferred, which is 2 in this case (since $Cd^{+2}$ and $Cu^{+2}$ both involve a 2-electron transfer).<br /><br />$Q$ is the reaction quotient, which can be calculated as:<br />$Q = \frac{[Cu^{+2}]}{[Cd^{+2}]} = \frac{0.45}{0.25} = 1.8$<br /><br />Substituting the values into the Nernst equation:<br />$E_{cell} = 0.740V - \frac{0.0592}{2} \log 1.8$<br />$E_{cell} = 0.740V - 0.0296 \log 1.8$<br />$E_{cell} = 0.740V - 0.0296 \times 0.255$<br />$E_{cell} = 0.740V - 0.0076$<br />$E_{cell} = 0.7324V$<br /><br />Therefore, the cell voltage is not $-0.421V$, so statement II is incorrect.<br /><br />III- The Free energy of the reaction is $-144.25kJ$.<br />To calculate the free energy of the reaction, we can use the formula:<br />$\Delta G = -nFE_{cell}$<br /><br />Where:<br />$n$ is the number of moles of electrons transferred, which is 2 in this case.<br />$F$ is the Faraday constant, which is approximately $96,485 C/mol$.<br />$E_{cell}$ is the cell voltage, which we calculated as $0.7324V$.<br /><br />Substituting the values into the formula:<br />$\Delta G = -2 \times 96,485 C/mol \times 0.7324V$<br />$\Delta G = -141,000 J/mol$<br />$\Delta G = -141 kJ/mol$<br /><br />Therefore, the free energy of the reaction is not $-144.25kJ$, so statement III is incorrect.<br /><br />Based on the analysis, none of the statements I, II, or III are correct. Therefore, the correct answer is D) Only III.