A cell can be prepared from copper and tin. What is the E° cellfor the cell that forms from the following half reactions?
Cu2+(aq) + 2e-Cu(s) E° = 0.35 V
Sn4+(aq) + 2e- Sn2+(aq) E° = 0.13 V
a. 0.48 V
b. 0.22 V
c. -0.48 V
d. -0.22 V
equation 1 written as is = E1 for
equation 2 reversed E for E2 = -0.13
E1 + E2 = 0.35 + (-0.13) = ??
thanks DR BOB, i did get that. they have given us an assignment for distant ed. students and we haven't even covered the topic yet. the assignment is due the day we start the electro chem topic! not good. thankyou
just one thing on this dr bob, i thought that it was Ecell= E*cathode-E*anode. Why do you have a + sign between the 2. E1+E2?
your way i get 0.35+(-0.13)=0.22, but my way i get 0.35-(-0.13)=0.48? which one?
a
Why did the cell go to therapy? Because it had some serious E° issues!
To find the E° cell for the cell formed from these half reactions, you simply add the E° values together. So, E° cell = E° Cu + E° Sn.
Plugging in the given values, E° cell = 0.35 V + 0.13 V = 0.48 V.
So the correct answer is a. 0.48 V.
To find the E°cell for the cell that forms from the given half reactions, you use the formula:
E°cell = E°cathode - E°anode
In this case, the copper (Cu) half reaction is reduction as it gains electrons, so it is the cathode. The tin (Sn) half reaction is oxidation as it loses electrons, so it is the anode.
From the given information, we have:
E°cathode (Cu) = 0.35 V
E°anode (Sn) = 0.13 V
Plugging these values into the formula, we get:
E°cell = 0.35 V - 0.13 V
E°cell = 0.22 V
Therefore, the E°cell for the cell that forms from these half reactions is 0.22 V.
The correct answer is option b. 0.22 V.