A concentration cell consists of two Sn/Sn2+ electrodes. The electrolyte in compartment A is 0.16 M Sn(NO3)2. The electrolyte in B is 0.86 M Sn(NO3)2.
Which electrode is the cathode?
What is the voltage of the cell? (Answer with two significant figures.)
Ecell = Eocell - (0.05916/n)*log(dilute/concd)
Eocell = 0
Substitute and solve for Ecell.
To determine which electrode is the cathode in a concentration cell, we need to look at the concentrations of the electrolytes in A and B.
In a concentration cell, the cathode is where reduction occurs, and the anode is where oxidation occurs. Reduction occurs at the electrode with the higher concentration of the species being reduced.
In this case, compartment B has a higher concentration of Sn2+ ions (0.86 M) compared to compartment A (0.16 M). Therefore, the Sn/Sn2+ electrode in compartment B is the cathode.
To find the voltage of the cell, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
where:
Ecell = cell potential
E°cell = standard cell potential (at 25°C, 298 K)
n = number of electrons transferred in the balanced reduction half-reaction
Q = reaction quotient
The standard cell potential for the Sn/Sn2+ half-reaction can be found in a standard electrode potential table. Let's assume that E°cell = 0.14 V.
Since we have Sn/Sn2+ on both sides of the cell, the balanced half-reaction for the cell is:
Sn2+ + 2e- -> Sn
The number of electrons transferred in this reaction is 2.
The reaction quotient (Q) can be determined using the concentrations of Sn2+ in compartments A and B:
Q = [Sn2+ in A] / [Sn2+ in B] = (0.16 M) / (0.86 M) = 0.186
Now, let's calculate the voltage of the cell using the Nernst equation:
Ecell = 0.14 V - (0.0592/2) * log(0.186)
Ecell ≈ 0.14 V - 0.0296 * (-0.729)
Ecell ≈ 0.14 V - 0.0215
Ecell ≈ 0.12 V
Therefore, the voltage of the cell is approximately 0.12 V (two significant figures).
To determine which electrode is the cathode in the concentration cell, we need to consider the concentrations of Sn2+ ions in each compartment. In a concentration cell, the electrode with the higher concentration of the species being reduced (in this case Sn2+) acts as the cathode.
In compartment A, the concentration of Sn2+ ions is 0.16 M, while in compartment B, the concentration of Sn2+ ions is 0.86 M. Since the concentration in compartment B is higher, the electrode in compartment B is the cathode.
Now, to calculate the voltage of the cell, we can use the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
Ecell is the cell potential
E°cell is the standard cell potential
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
n is the number of electrons involved in the reaction
F is the Faraday constant (96,485 C/mol)
Q is the reaction quotient (the ratio of products to reactants)
In this case, the reaction at each electrode is:
Cathode: Sn2+ (higher concentration) + 2e- -> Sn(s)
Anode: Sn(s) -> Sn2+ (lower concentration) + 2e-
Since the reaction in this concentration cell is the same at both electrodes, n = 2.
To find E°cell, we need the standard reduction potential of the Sn2+/Sn half-reaction. The standard reduction potential for this reaction is -0.136 V.
Now, let's substitute the values into the Nernst equation.
Ecell = -0.136 V - (8.314 J/(mol·K) * T/(2 * 96,485 C/mol) * ln(Q)
Since we don't have information on the temperature or the actual values for the reaction quotient, we cannot calculate the exact voltage of the cell. However, using the Nernst equation, you can substitute the known values once you have the temperature and concentrations to calculate the cell voltage. Remember to round the answer to two significant figures.