Question

For the following cell reaction, write the half reactions occurring at the anode and cathode and determine the electromotive force (EMF) of the cell:
\( 2Ag^+ + Cd \rightarrow 2Ag + Cd^{2+} \)
(Given, \( E^{\circ}_{Ag^+/Ag} = +0.80\ V \) and \( E^{\circ}_{Cd^{2+}/Cd} = -0.40\ V \))

Hint:

Cd is oxidised (anode), Ag+ is reduced (cathode); \( E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \).

Answer 1

Step 1: Decide which species is oxidised and which is reduced.
In the cell reaction \( 2Ag^+ + Cd \rightarrow 2Ag + Cd^{2+} \), cadmium metal loses electrons (its oxidation state goes from 0 to +2), so it is oxidised. Silver ions gain electrons (from +1 to 0), so they are reduced.

Step 2: Write the half reactions.
Oxidation occurs at the anode:
\[ Cd \rightarrow Cd^{2+} + 2e^- \quad (anode) \]
Reduction occurs at the cathode:
\[ 2Ag^+ + 2e^- \rightarrow 2Ag \quad (cathode) \]

Step 3: Formula for cell EMF.
\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]
Here the cathode is the silver electrode and the anode is the cadmium electrode.

Step 4: Substitute the given values.
\[ E^{\circ}_{cell} = E^{\circ}_{Ag^+/Ag} - E^{\circ}_{Cd^{2+}/Cd} = (+0.80) - (-0.40) \]

Step 5: Arithmetic.
\[ E^{\circ}_{cell} = 0.80 + 0.40 = 1.20\ V \]
The positive value confirms the reaction is spontaneous.
\[\boxed{E^{\circ}_{cell} = +1.20\ V}\]