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Function of salt bridge in electrochemical cells?
Salt bridges are usually immersed in a solution of some ionic compound.
If there was no salt bridge. That means that electrons will have a hard time leaving the anode due to an excess of the positive ion that is produced from the oxidation.
Is that basically the function of the salt bridge?
And for the reduction potential it seems that the excess negative charge would repel the electrons
This is for galvanic cells
6 Answers
- 1 decade agoFavorite Answer
IThe salt bridge maintains neutrality in the charges (giving and taking electrons) between the anode and cathode. This is pretty much how electrolytes get from one side of a unit to the other side.
Reduction potentials are tricky. If you have a positive reduction potential, most likely you are dealing with a strong oxidizing agent (reduces a reactant).
If you mean the excess electrons on an ion with a negative, then on the product side a charge of 0, you would have added electrons to the reactant side to make the oxidation charge on a certain atom of a compound equal to its oxidation charge on the product side; the overall negative charge does not mean it will repel electrons (because electrons are a reactant).
Do a few tricky redox problems in acidic and basic solutions and this stuff is a piece of cake.
- 1 decade ago
You are correct. A salt bridge allows ion flow to maintain a balance of charge between reaction cells
- Zor PrimeLv 71 decade ago
Lancenigo di Villorba (TV), Italy
You know, Electrodical Cell works everytime you get two electrical
half-elements named Cathode and Anode, e.g. the former one hosting (+) Polarity or Higher Electrical Potential.
You know, once you built two electrodes, you have to connect it by metallic wire : metallic wire will lead electrons or Electrical Currents flowing between electrodes until there is a satisfacting Electromotive Force warranted by Electrical Potential.
These statements come from NERNST LAW about electrodes and OHM LAW about electrical circuit.
ELECTRICAL DOUBLE LAYER
Really, there is a fault since this ignores the role of Electrical Double Layers.
Every electrode may result a metallic terminal touching liquid medium.
So, a short-reached equilibrium determine the concentration of Metal Ions standing-in-front of metallic surface ; these ions attracted counterions as electrically negative ions which build a second electrical layer onto metal ion's one, e.g. the Double Layer.
You understand that to support a satisfacting electrical current along metallic wire THERE SHOULD BE AN ELECTRICALLY EQUIVALENT RELEASE OF METALLIC IONs IN LIQUID MEDIA BUT THIS LATTER HAS BEEN BRAKED BY Double Layer WHICH OPPOSE.
SALT BRIDGE
In order to breakdown this resistance, YOU HAVE TO ASSURE ELECTRICAL CHARGE TRANSPORT FROM Metallic Plates, FROM Double Layers itselves : THIS CHARGE TRANSPORT MEANS IONIC JOINT AND IT HAS BEEN ASSURED BY Salt Bridge. SALT BRIDGE has to get its Positive Ions into Cathodic Vat whilst it leaves Negative Ions into Anodic Vat. SALT BRIDGE is a useful device built by a Ionic Buffer inside a Pipe. This device serves to assures Ionic Exchange among electrodical vats.
I hope this helps you.
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- Anonymous5 years ago
hello dear iam sharath 18 b,sc ican give usimple and sure short ans. as for a voltoic cell we need acathode and anode must ! and they should be dissolved intheir respective like sulphate and nitarate solutions. as we know here cathode means where reductiom takes place amd anode where oxidation takes plase with this proprty omly we r doing electro plating techinique,and coming to salt bridge itis passage which easily allows ions to flowwhich is madeof chemical composition, as here due to flowthat means the exchange of ions only we get current through which apotentiol is generatedas weknow electrodepotentiol is E right - Eleft AND (R) IS CATHODE AND (L)IS ANODEIF U WANT TO CHAT CHAT MORE WITH UCAN PLEASE AS IAM FEELIMG LONE NOWA DAYS IGOT 86%IN CBSE IN +2MY ID IS sharathh_123 PLEASEEEEEEEEEEEE BYE
- biire2uLv 71 decade ago
You probably already saw this, but just in case you haven't, take a look at this explanation under the paragraph, "Uses":