[johnnyjohn993123]

If  the cell emf of a Zn-Cu cell is 0.80 V when the conc of Zn²⁺ is 2.0M, what is the conc. of Cu²⁺?

how will I know whether what species underwent reduction or oxidation ? and what is emf? is It  Ecell in Nernst's Equation ?

[mjc123]

Yes it is. Ecell is the emf of the cell when it is in equilibrium. Neither reduction nor oxidation is taking place. (Or rather, since the equilibrium is dynamic, they are both taking place at the same rate, with no net reaction.) There is some Zn²⁺ present and some Cu²⁺ - how much of each determines the value of Ecell. If you apply a more positive or more negative voltage, you can make the reaction go in one direction or the other.

The answer to your other question is basically the same.

[johnnyjohn993123]

Yes it is. Ecell is the emf of the cell when it is in equilibrium. Neither reduction nor oxidation is taking place. (Or rather, since the equilibrium is dynamic, they are both taking place at the same rate, with no net reaction.) There is some Zn²⁺ present and some Cu²⁺ - how much of each determines the value of Ecell. If you apply a more positive or more negative voltage, you can make the reaction go in one direction or the other.

The answer to your other question is basically the same.

how may I able to know what to plug in E° if I dont know what species oxidized or reduced?
that is if use the equation Ecell= E°- RT/Fn ln x/y to determine what the conc. of Cu²⁺ and whether it is x or y from the equation ?

[mjc123]

You know what species are present - Zn, Zn²⁺, Cu and Cu²⁺. Work out E for the Zn²⁺/Zn half cell, and E for the Cu²⁺/Cu half cell. Subtract them.

[johnnyjohn993123]

i go directly using this equation
 E =E° - RT/nF ln x/y

hypothesis one:
Zn°  Zn²⁺ +2e^-
Cu²⁺ +2e^-    Cu°

E°= 0.34 V -(-0.76)
E°= 1.1 V
USING THIS
 E =E° - RT/nF ln x/y
0.80 v= 1.1V - RT/2 F  ln (2.0)/[Cu²⁺]   ψAlgebraψ
[Cu²⁺] = 1.45x10^-10 M

hypothesis two:
Zn²⁺ +2e^-  Zn⁰
Cu°   Cu²⁺ +2e^-
E°= (-0.76)-0.34 V
E° = -1.1 V

USING THIS
 E =E° - RT/nF ln x/y
0.80 v= -1.1V - RT/2 F  ln [Cu²⁺] /(2.0)  ψAlgebraψ
 [Cu²⁺]= 1.22x10^-64 M

So what is the right answer between the two now? this confuses me like alot

[mjc123]

The first way is correct.
There is a slight ambiguity because we are not told the polarity of the cell. If we work out E° for a cell where the concentrations are 1M, we take the half-cell E°s and subtract the less positive from the more positive, so that E°_cell is positive. Maintaining the convention, in the case of hypothesis 2 the concentrations are so changed that the polarity is reversed, and E_cell would be -0.8V. But somebody just measuring the voltage might say it was 0.8V, without specifying the polarity.
But you can tell that hypothesis 2 is physically unrealistic. To produce a change of ±1.9V just by changing the concentrations requires a huge concentration change. Think about it. If you have, say, 1L of a solution 10^-64 M in Cu²⁺, how many copper atoms would it contain?