Question

Calculate $E^0_{cell}$ for $Cd(s) | Cd^{++}(1M) | | Ag^{+}(1M) | Ag(s)$. Given $E^0_{Cd} = -0.403 V$; $E^0_{Ag} = 0.799 V$.

• A. 1.202 V
• B. -1.202 V
• C. 0.396 V
• D. -0.396 V

Hint:

In a galvanic (voltaic) cell, the species with the more positive (or less negative) standard reduction potential ($E^0$) acts as the cathode (undergoes reduction), and the species with the more negative (or less positive) $E^0$ acts as the anode (undergoes oxidation). The standard cell potential is calculated as $E^0_{cell} = E^0_{\text{cathode - E^0_{\text{anode$.

Answer 1

The Correct Option is A

Step 1: Identify anode and cathode.\ The cell notation $Cd(s) | Cd^{++}(1M) | | Ag^{+}(1M) | Ag(s)$ indicates that Cadmium ($Cd$) is the anode (oxidation occurs here) and Silver ($Ag$) is the cathode (reduction occurs here).\ \ Given standard reduction potentials:\ $E^0_{Cd} = -0.403\text{ V}$ (for $Cd^{2+} + 2e^- \to Cd$)\ $E^0_{Ag} = 0.799\text{ V}$ (for $Ag^{+} + e^- \to Ag$)\ \ Since $Cd$ has a more negative standard reduction potential than $Ag$, $Cd$ will be oxidized (anode) and $Ag^+$ will be reduced (cathode).\ \ Oxidation at anode: $Cd(s) \to Cd^{2+}(aq) + 2e^-$ \ Reduction at cathode: $Ag^{+}(aq) + e^- \to Ag(s)$ \
Step 2: Apply the formula for standard cell potential.\ The standard cell potential ($E^0_{cell}$) is calculated using the formula:\ \[ E^0_{cell} = E^0_{\text{cathode - E^0_{\text{anode \]\
Step 3: Substitute values and calculate.\ Substitute the given standard reduction potentials into the formula:\ \[ E^0_{cell} = E^0_{Ag} - E^0_{Cd} \]\ \[ E^0_{cell} = (0.799\text{ V}) - (-0.403\text{ V}) \]\ \[ E^0_{cell} = 0.799\text{ V} + 0.403\text{ V} \]\ \[ E^0_{cell} = 1.202\text{ V} \]\ Therefore, the standard cell potential for the given cell is $1.202\text{ V}$.