Find the Emf for the cell
Cr(gr) | Sn4+(aq, .059 M), Sn2+ (aq, .059 M) || Fe3+ (aq, .15 M), Fe2+ (aq, .015 M) | Pt(s) if the standard cell emf is .77 V at the cathode and .15 V at the anode. Answer in Units of V. If would be nice if you could show me step by step.
To find the emf (electromotive force) for the cell, we need to use the Nernst equation. The Nernst equation allows us to calculate the cell potential under non-standard conditions.
The Nernst equation is as follows:
E = E° - (RT / nF) × ln(Q)
Where:
E = Cell potential under non-standard conditions
E° = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
n = Number of moles of electrons transferred in the balanced chemical equation
F = Faraday's constant (96,485 C/mol)
ln = Natural logarithm
Q = Reaction quotient
Now, let's proceed step by step to find the emf for the given cell:
Step 1: Write the balanced redox reaction for the given cell.
Cr(gr) + Sn4+(aq) + 4e- → Cr3+(aq) + Sn2+(aq) (Cathode)
Fe3+(aq) + e- → Fe2+(aq) (Anode)
Step 2: Determine the number of moles of electrons transferred.
From the balanced reactions, we can see that 4 moles of electrons are transferred in the cathode reaction and 1 mole of electrons in the anode reaction.
Step 3: Calculate the reaction quotient (Q) for the cell.
The reaction quotient (Q) is the concentration of the products raised to their stoichiometric coefficients divided by the concentration of the reactants raised to their stoichiometric coefficients.
Q = [Cr3+(aq)][Sn2+(aq)] / [Sn4+(aq)][Fe2+(aq)]
For the given concentration values:
Q = (0.059)(0.059) / (0.15)(0.015)
Step 4: Calculate the emf (E) using the Nernst equation.
Substitute the known values into the Nernst equation:
E = E° - (RT / nF) × ln(Q)
E = (0.77 V) - [(8.314 J/(mol·K)) × (298 K) / (4 × 96,485 C/mol) × ln((0.059)(0.059) / (0.15)(0.015))]
E = 0.77 V - 0.0572 V
Step 5: Calculate the final emf for the cell.
E = 0.7128 V
Therefore, the emf for the given cell is approximately 0.7128 V.