Question

question 3 (5 points)
listen
cu+2 + 2e- → cu e° = 0.52 v
ag+1 + 1e- → ag e° = 0.80 v
using the two cell reduction potentials shown for their corresponding reaction, calculate the cell potential for a voltaic cell made from these two metals.
a) -0.28 v
b) 1.32 v
c) 0.28 v
d) -1.32 v

Explanation:

Step1: Identify cathode and anode

In a voltaic cell, the reaction with the higher reduction potential occurs at the cathode (reduction), and the one with the lower reduction potential occurs at the anode (oxidation, so reverse the reaction). Here, \( E^{\circ}_{\text{Ag}^+/\text{Ag}} = 0.80\ \text{V} \) (higher) and \( E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = 0.52\ \text{V} \) (lower). So, reduction at cathode: \( \text{Ag}^+ + e^-
ightarrow \text{Ag} \) (\( E^{\circ}_{\text{cathode}} = 0.80\ \text{V} \)), oxidation at anode: \( \text{Cu}
ightarrow \text{Cu}^{2+} + 2e^- \) (reverse of \( \text{Cu}^{2+} + 2e^-
ightarrow \text{Cu} \), so \( E^{\circ}_{\text{anode (oxidation)}} = - 0.52\ \text{V} \), but when calculating cell potential, we use \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode (reduction)}} - E^{\circ}_{\text{anode (reduction)}} \)).

Step2: Calculate cell potential

The formula for cell potential of a voltaic cell is \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode (reduction)}} - E^{\circ}_{\text{anode (reduction)}} \). Here, \( E^{\circ}_{\text{cathode}} = 0.80\ \text{V} \) (reduction of \( \text{Ag}^+ \)), \( E^{\circ}_{\text{anode (reduction)}} = 0.52\ \text{V} \) (reduction of \( \text{Cu}^{2+} \)). So, \( E^{\circ}_{\text{cell}} = 0.80 - 0.52 = 0.28\ \text{V} \).

Answer:

C) 0.28 V