Question

Consider the cell described below at 273 K: Sn | Sn2+ (0.835 M) || Pb2+ (0.961...

Consider the cell described below at 273 K:

Sn | Sn2+ (0.835 M) || Pb2+ (0.961 M) | Pb

Given EoPb2+→Pb = -0.131 V, EoSn2+→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.385 mol/L.

Science   Chemistry   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Consider the cell described below at 257 K: Sn | Sn2+ (0.811 M) || Pb2+ (0.955...
Consider the cell described below at 257 K: Sn | Sn2+ (0.811 M) || Pb2+ (0.955 M) | Pb Given EoPb2+→Pb = -0.131 V, EoSn2+→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.399 mol/L.
Consider the cell described below at 255 K: Sn | Sn2+ (0.969 M) || Pb2+ (0.941...
Consider the cell described below at 255 K: Sn | Sn2+ (0.969 M) || Pb2+ (0.941 M) | Pb Given EoPb2+→Pb = -0.131 V, EoSn2+→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.323 mol/L.
Consider the cell described below at 275 K: Sn | Sn2+ (0.775 M) || Pb2+ (0.931...
Consider the cell described below at 275 K: Sn | Sn2+ (0.775 M) || Pb2+ (0.931 M) | Pb Given EoPb2+→Pb = -0.131 V, EoSn2+→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.381 mol/L. Tries 0/45
Consider the cell described below at 261 K: Sn | Sn2+ (0.777 M) || Pb2+ (0.957...
Consider the cell described below at 261 K: Sn | Sn2+ (0.777 M) || Pb2+ (0.957 M) | Pb Given EoPb2+→Pb = -0.131 V, EoSn2+→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.379 mol/L. I got an E standard of .17 I then used the nernst equation using concentrations that I got from setting up an ice table. E=.17-((.059/2) ln(1.336/.398)) = .1343 V Does anyone know...
Consider the cell described below at 273 K: Sn | Sn2+ (1.45 M) || Fe3+ (2.33...
Consider the cell described below at 273 K: Sn | Sn2+ (1.45 M) || Fe3+ (2.33 M) | Fe Given the standard reduction potentials found on the sheet attached to the exam, calculate the cell potential after the reaction has operated long enough for the [Fe3+] to have changed by 1.242 M
Consider the cell described below at 263 K: Fe | Fe2+ (0.865 M) || Cd2+ (0.941...
Consider the cell described below at 263 K: Fe | Fe2+ (0.865 M) || Cd2+ (0.941 M) | Cd Given EoCd2+→Cd = -0.403 V, EoFe2+→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.391 mol/L.
Consider the cell described below at 287 K: Fe | Fe2+ (0.829 M) || Cd2+ (0.953...
Consider the cell described below at 287 K: Fe | Fe2+ (0.829 M) || Cd2+ (0.953 M) | Cd Given EoCd2+→Cd = -0.403 V, EoFe2+→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.301 mol/L.
Consider the cell described below at 285 K: Fe | Fe2+ (0.919 M) || Cd2+ (0.965...
Consider the cell described below at 285 K: Fe | Fe2+ (0.919 M) || Cd2+ (0.965 M) | Cd Given EoCd2+→Cd = -0.403 V, EoFe2+→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.391 mol/L.
Consider the cell described below at 259 K: Fe | Fe2+ (0.853 M) || Cd2+ (0.907...
Consider the cell described below at 259 K: Fe | Fe2+ (0.853 M) || Cd2+ (0.907 M) | Cd Given EoCd2+→Cd = -0.403 V, EoFe2+→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.321 mol/L.
A voltaic cell is constructed so its based on the reaction Sn2+(aq)+Pb(s)=Sn(s)+Pb2+(aq) a. If the concentration...
A voltaic cell is constructed so its based on the reaction Sn2+(aq)+Pb(s)=Sn(s)+Pb2+(aq) a. If the concentration of Sn2+ in the cathode compartment is 1.00 M and the cell generates an emf of 0.15 V , what is the concentration of Pb2+ in the anode compartment? b. If the anode compartment contains [SO2−4]= 1.20 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT