Negative emf

[determined daisy]

What does Negative emf indicates?What is concept behind Negative emf ?I am not talking about Negative induced emf,I am talking in context of electrochemical cell.

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[Cawolf]

The cell is non-spontaneous and will only run if a voltage equal or greater to the -emf is applied.

In other words, it will only run electrolytically.

 

[determined daisy]

I am asking in the sense that
E°cell = E°red (cathode) – E°red (anode)
so can you explain negative emf in terms of his formula,

 

[EParker37]

I am asking in the sense that
E°cell = E°red (cathode) – E°red (anode)
so can you explain negative emf in terms of his formula,

Cawolf has it correct. Essentially if your cell potential is negative, the reaction will not occur until you apply a voltage greater than the cell potential, but this is going to run in the reverse of a galvanic cell, and will deposit onto the anode and oxidize the cathode. An example would be electroplating.

 

[Labrat07]

I am asking in the sense that
E°cell = E°red (cathode) – E°red (anode)
so can you explain negative emf in terms of his formula,

I think this is what you're looking for:

Look at this equation. Know spontaneous = positive E cell. We have two two electrodes
AN OX RED CAT.
Oxidation at anode and reduction at cathode. Electron go from anode to cathode. This addition/substraction must be positive for galnanic to happen. If E cell is negative, it need something to push. For example: current (electrolysis)

I'd use

E°cell = E°red (cathode) + E°red (anode) though. Easier to understand and use.

 

[NextStepTutor_1]

In that equation, there are possibilities that the E of reduction and E of oxidation can be negative or positive. So it really depends on what you want to reduce and what you want to oxidize (and how favorable, or positive, the E of the situation is).
Definitely know the Red Cat, An Ox mnemonic, but more importantly know how to read a set of reactions, quickly realize if they are oxidation or reduction rxns, flip any according to the question, and add up the E values. This is the type of question that appeared at a very high frequency on the old test.