Problem: Consider a voltaic cell based on the half-cells:Ag+(aq) + e- → Ag(s) E° = +0.80 VSn2+(aq) + 2 e- → Sn(s) E° = -0.14 VIdentify the cathode and give the cell voltage under standard conditions:a. Sn; E°cell = +0.94b. Sn2+; E°cell = +0.94c. Sn; E°cell = -0.94d. Ag; E°cell = +0.94e. Ag; E°cell = +0.94

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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)

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Problem Details

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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