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: Rank the electrodes by reduction potential.
The silver couple has the higher (more positive) standard reduction potential, \(+0.80\) V, so it prefers to be reduced and becomes the cathode. The cadmium couple, at \(-0.40\) V, is forced to undergo oxidation and becomes the anode.

Step 2: Electrode half reactions.
Cathode (reduction): \(2Ag^+ + 2e^- \rightarrow 2Ag\).
Anode (oxidation): \(Cd \rightarrow Cd^{2+} + 2e^-\).
Both involve two electrons, so they add directly to the given overall equation.

Step 3: Use the sum of potentials method.
Write \(E^{\circ}_{cell}\) as the reduction potential of the cathode plus the oxidation potential of the anode. The oxidation potential of cadmium is the negative of its reduction potential, that is \(+0.40\) V.
\(E^{\circ}_{cell} = E^{\circ}_{red}(Ag^+/Ag) + E^{\circ}_{ox}(Cd/Cd^{2+}) = 0.80 + 0.40\).

Step 4: Result.
\[\boxed{E^{\circ}_{cell} = 1.20\ V}\]
Being positive, it also tells us \(\Delta G^{\circ} = -nFE^{\circ}_{cell}\) is negative, so the cell works spontaneously.