We’re asked to determine the Ecell and the cathode for the half-reactions:
Ag+(aq) + e-→ Ag(s) E° = +0.80 V
Sn2+(aq) + 2 e-→ Sn(s) E° = -0.14 V
For this problem, we have to determine the anode and cathode reactions:
Loss of Electron is an Oxidation process undergone by the Reducing Agent (LEORA). In the Anode, Oxidation happens (An Ox)
Consider a voltaic cell based on the half-cells:
Ag+(aq) + e- → Ag(s) E° = +0.80 V
Sn2+(aq) + 2 e- → Sn(s) E° = -0.14 V
Identify the cathode and give the cell voltage under standard conditions:
a. Sn; E°cell = +0.94
b. Sn2+; E°cell = +0.94
c. Sn; E°cell = -0.94
d. Ag; E°cell = +0.94
e. Ag; E°cell = +0.94
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Based on our data, we think this problem is relevant for Professor Young's class at UCSD.