3. For the reaction between Cu and Sn the potentials are given below;
Cu2+(aq) +2e- -----------> Cu(s) Eºcathode = 0.34 (half-cell potential)
Sn(s) -----------> Sn2+(aq) + 2 e- Eºanode = - 0.14 (half-cell potential)
Calculate the E.M.F (Voltage) of the cell
Calculate pH given [H+] = 1.4 x 10-5 M,
Find [H+] if pH = 8.5
EMF = ? Add the oxidation half to the reduction half.
Use pH = -log (H^+) for the other two.
To calculate the electromotive force (EMF) or voltage of the cell, you can use the equation:
EMF = Eºcathode - Eºanode
EMF = 0.34 V - (-0.14 V)
EMF = 0.48 V
Therefore, the voltage of the cell is 0.48 V.
To calculate pH given [H+] = 1.4 x 10-5 M, you can use the equation:
pH = -log[H+]
pH = -log(1.4 x 10^-5)
pH = -log(1.4) - log(10^-5)
pH = -0.146 - (-5)
pH = -0.146 + 5
pH = 4.854
Therefore, the pH is 4.854.
To find [H+] if pH = 8.5, you can use the equation:
[H+] = 10^(-pH)
[H+] = 10^(-8.5)
[H+] = 3.162 x 10^(-9) M
Therefore, [H+] is 3.162 x 10^(-9) M.