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^aCu^n/Cu=0.337V;E^circ Ca^n/Ca^n-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) 1 and II B) I and III C II and III D) Only 1 E) 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 />The standard reduction potentials are:<br />\[ E^{\circ} (Cu^{+2}/Cu) = 0.337V \]<br />\[ E^{\circ} (Cd^{+2}/Cd) = -0.403V \]<br /><br />Let's break down each statement:<br /><br />I. **The cell type is an electrolytic cell:**<br /> - This statement is incorrect. The cell notation indicates it is a galvanic (or voltaic) cell because it involves a spontaneous redox reaction where electrons flow from the anode (Cd) to the cathode (Cu).<br /><br />II. **Cell Voltage is -0.421V:**<br /> - To find the cell voltage (E_cell), we use the Nernst equation for each half-reaction and then calculate the overall cell potential.<br /> - For the reduction half-reaction (Cu^+2 + 2e^- → Cu):<br /> \[ E_{Cu} = E^{\circ}_{Cu} + 0.0592 \log \left( \frac{[Cu^{+2}]}{[Cu]} \right) \]<br /> \[ E_{Cu} = 0.337V + 0.0592 \log \left( \frac{0.45}{1} \right) \]<br /> \[ E_{Cu} = 0.337V + 0.0592 \times (-0.356) \]<br /> \[ E_{Cu} = 0.337V - 0.0212V \]<br /> \[ E_{Cu} = 0.3158V \]<br /> - For the oxidation half-reaction (Cd → Cd^+2 + 2e^-):<br /> \[ E_{Cd} = E^{\circ}_{Cd} + 0.0592 \log \left( \frac{[Cd^{+2}]}{[Cd]} \right) \]<br /> \[ E_{Cd} = -0.403V + 0.0592 \log \left( \frac{0.25}{1} \right) \]<br /> \[ E_{Cd} = -0.403V + 0.0592 \times (-0.602) \]<br /> \[ E_{Cd} = -0.403V - 0.0357V \]<br /> \[ E_{Cd} = -0.4387V \]<br /> - The overall cell voltage (E_cell) is the difference between the cathode and anode potentials:<br /> \[ E_{cell} = E_{Cu} - E_{Cd} \]<br /> \[ E_{cell} = 0.3158V - (-0.4387V) \]<br /> \[ E_{cell} = 0.3158V + 0.4387V \]<br /> \[ E_{cell} = 0.7545V \]<br /> - This statement is incorrect because the calculated cell voltage is 0.7545V, not -0.421V.<br /><br />III. **The Free energy of the reaction is -144.25kJ:**<br /> - The free energy change (ΔG) can be calculated using the formula:<br /> \[ \Delta G = -nFE_{cell} \]<br /> where \( n \) is the number of moles of electrons transferred (2 moles in this case, since each Cd atom loses 2 electrons and each Cu^+2 ion gains 2 electrons).<br /> \[ \Delta G = -2 \times 96485 \times 0.7545 \]<br /> \[ \Delta G = -145,678.7 J \]<br /> \[ \Delta G = -145.7 kJ \]<br /> - This statement is approximately correct, as the calculated free energy change is -145.7 kJ, which is close to -144.25 kJ.<br /><br />Therefore, the correct answer is:<br /><br />E) Only III