In this redox reaction, Ti+3 is oxidized to Ti+4, and Sn+2 is reduced to Sn+4:
Ti+3 + Sn+2 → Ti+4 + Sn+4
The balanced equation shows that one mole of Ti+3 reacts with one mole of Sn+2. We can use this to calculate the number of moles of Sn+2 required to react completely with the Ti+3 in the solution:
n(Sn+2) = (0.30 mol/L) x (0.040 L) = 0.012 mol
The volume of Sn+2 required to react with the Ti+3 can be calculated using the molarity and the number of moles:
V(Sn+2) = n(Sn+2) / [Sn+2] = 0.012 mol / 0.75 mol/L = 0.016 L = 16 mL
At the equivalence point, all of the Ti+3 has reacted with the Sn+2, so the resulting solution contains only Ti+4 and Sn+4. The potential at the equivalence point can be calculated using the Nernst equation:
E = E° - (RT/nF) x ln(Q)
where E° is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is the Faraday constant, and Q is the reaction quotient.
The standard cell potential can be calculated using the reduction potentials for the half-reactions:
Ti+4 + 2 e- → Ti+3 E° = -0.15 V
Sn+4 + 2 e- → Sn+2 E° = 0.15 V
The overall cell potential is the sum of the reduction potentials:
E°cell = E°reduction (cathode) - E°reduction (anode) = 0.15 V + 0.15 V = 0.30 V
At the equivalence point, the reaction quotient Q is equal to the equilibrium constant K:
K = [Ti+4] / [Sn+4]
The concentrations of Ti+4 and Sn+4 can be calculated from the number of moles and the total volume of the solution:
n(Ti+4) = n(Sn+2) = 0.012 mol
V(total) = 0.040 L + 0.016 L = 0.056