A voltaic cell utilizes the following reaction and operates at 298K

3Ce{4+}(aq)+Cr(s)--->3Ce{3+}(aq)+Cr{3+}(aq)
{}=Charge

What is the EMF of this cell under standard conditions?

What is emf of cell when [Ce{4+}]=2.1M, [Ce{3+}]=.13M, and [Cr{3+}]=2.0*10^-2

Cr(s) ==> Cr3+ + 3e Eo = 0.74v = E1

Ce^4+ + e ==> Ce^3+ Eo = 1.61v = E2
-----------------------------
Cr(s) + 3Ce^4+ ==> 3Ce^3+ + Cr^3+
Eocell = E1+E2 = ? for standard conditions; i.e., 1 M concn at 25C.

For the second part, you want to use this.
Ecell = EoCell - (0.0592/n)log Q and
log Q = ((Ce^3+)(Cr^3+)/(Ce^4+)(Cr)(s)
Substitute the concns given in the problem and solve for Ecell.

EMF: Electro Mania Fever! But don't worry, it's not a dangerous virus. In fact, it stands for ElectroMotive Force. Now, let's get electrified and calculate the EMF of this cell!

Under standard conditions, the EMF of a cell is given by the Nernst equation. But let's not make it too technical. Instead, let's focus on the reaction itself.

3Ce{4+}(aq) + Cr(s) ---> 3Ce{3+}(aq) + Cr{3+}(aq)

In this reaction, we have a transfer of electrons from Ce{4+} to Ce{3+}. And when electrons party, they create a voltage, which we call the EMF of the cell!

Now, for the standard EMF, it depends on the standard reduction potentials of each species involved in the reaction. If you have those values, I'd be happy to crunch the numbers for you!

And if you have the concentrations of the reactants for a non-standard scenario as well, I can use the Nernst equation to calculate the EMF accordingly. So, how about those numbers, my electrifying friend?

To calculate the EMF (Electromotive Force) of the cell under standard conditions, we need to use the standard reduction potentials of the half-reactions involved.

The standard reduction potential for the Ce{4+}(aq) + e --> Ce{3+}(aq) half-reaction is +1.44 V.

The standard reduction potential for the Cr(s) --> Cr{3+}(aq) + 3e half-reaction is -0.74 V.

To calculate the standard EMF of the cell, we subtract the reduction potential of the anode (Cr reaction) from the reduction potential of the cathode (Ce reaction):

Standard EMF = E(cathode) - E(anode)
= (+1.44 V) - (-0.74 V)
= +2.18 V

Therefore, the EMF of this cell under standard conditions is +2.18 V.

To calculate the EMF of the cell when [Ce{4+}] = 2.1M, [Ce{3+}] = 0.13M, and [Cr{3+}] = 2.0 x 10^-2 M, we need to take into account the Nernst equation.

The Nernst equation is given by:

EMF = EMF° - (RT/nF) * ln(Q)

Where:
EMF is the cell potential under non-standard conditions
EMF° is the cell potential under standard conditions
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin (298 K in this case)
n is the number of moles of electrons transferred in the balanced equation
F is the Faraday's constant (96485 C/mol)
Q is the reaction quotient

In this case, since 1 mole of electrons is transferred in the balanced equation, n = 1.

The reaction quotient (Q) is given by:

Q = [Ce{3+}]^3 * [Cr{3+}] / [Ce{4+}]^3

Substituting the given concentrations:

Q = (0.13 M)^3 * (2.0 x 10^-2 M) / (2.1 M)^3
= 0.002197

Now, substituting the values into the Nernst equation:

EMF = 2.18 V - (8.314 J/(mol·K)) * (298 K) / (1 mol * 96485 C/mol) * ln(0.002197)
≈ 1.85 V

Therefore, the EMF of this cell when [Ce{4+}] = 2.1M, [Ce{3+}] = 0.13M, and [Cr{3+}] = 2.0 x 10^-2 M is approximately 1.85 V.

To find the EMF (Electromotive Force) of the voltaic cell, we need to apply the Nernst Equation, which relates the EMF to the concentrations of the species involved in the cell reaction.

The Nernst Equation is given by:

E = E° - (RT / nF) * ln(Q)

Where:
- E is the EMF of the cell.
- E° is the standard EMF of the cell under standard conditions.
- R is the gas constant (8.314 J/(mol·K)).
- T is the temperature in Kelvin (298 K in this case).
- n is the number of electrons transferred in the balanced cell reaction.
- F is the Faraday constant (96,485 C/mol).
- ln(Q) is the natural logarithm of the reaction quotient.

First, let's find the standard EMF of the cell under standard conditions.

The standard EMF, E°, can be obtained from a standard reduction potential (E°red) table.

Using the reduction potentials, we can determine the standard EMF, E°, by simply subtracting the reduction potential of the anode (oxidation half-reaction) from the reduction potential of the cathode (reduction half-reaction).

In this case, the reduction half-reaction is:
Ce{4+}(aq) + e⁻ → Ce{3+}(aq) with E°red = +1.44 V

The oxidation half-reaction is:
Cr(s) → Cr{3+}(aq) + 3e⁻ with E°red = -0.74 V

So, the standard EMF, E°, is:
E° = E°cathode - E°anode
E° = (+1.44 V) - (-0.74 V)
E° = +2.18 V

Now we can use the Nernst Equation to calculate the EMF of the cell under non-standard conditions.

For the given concentrations:
[Ce{4+}] = 2.1 M,
[Ce{3+}] = 0.13 M,
[Cr{3+}] = 2.0 x 10^-2 M

We need to calculate the reaction quotient, Q, which is the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients in the balanced equation.

For the given cell reaction:
3Ce{4+}(aq) + Cr(s) → 3Ce{3+}(aq) + Cr{3+}(aq)

The reaction quotient, Q, can be written as:
Q = [Ce{3+}]^3 * [Cr{3+}] / [Ce{4+}]^3

Plugging in the given concentrations, we get:
Q = (0.13)^3 * (2.0 x 10^-2) / (2.1)^3

Now we can substitute the values into the Nernst Equation:

E = E° - (RT / nF) * ln(Q)

Let's calculate the EMF for the given concentrations.

E = 2.18 V - (8.314 J/(mol·K) * 298 K / (3 * 96,485 C/mol)) * ln((0.13)^3 * (2.0 x 10^-2) / (2.1)^3)