alleryFor a galvanic cell, the standard cell EMF is given by:
A$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} + E^{\circ}_{\text{anode}}$
B$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{anode}} - E^{\circ}_{\text{cathode}}$
C$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$
D$E^{\circ}_{\text{cell}} = -2 E^{\circ}_{\text{anode}}$
Answer & Solution
Correct answer: C. $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$
$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$, where both potentials are listed as standard *reduction* potentials.
For a spontaneous galvanic cell, $E^{\circ}_{\text{cell}} > 0$, which means the cathode has the higher reduction potential. This is just an algebraic restatement of "reduction happens at the cathode".
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