A voltaic cell with Mn/Mn2+ and Cd/Cd2+ half-cells has the following initial concentrations:
[Mn2+]= 0.090 M; [Cd2+ ] = 0.060 M.
What is E
cell
when [Cd2+ ] reaches 0.035 M?
To calculate the cell potential when [Cd2+] reaches 0.035 M, we first need to write the balanced redox reaction and the cell notation.
Balanced redox reaction:
Mn2+ + 2e- -> Mn (s)
Cd2+ + 2e- -> Cd (s)
Cell notation:
Mn(s) | Mn2+ (0.090 M) || Cd2+ (0.060 M) | Cd(s)
Now let's calculate the standard reduction potentials for both half-reactions:
E°(Mn2+/Mn) = -1.18 V
E°(Cd2+/Cd) = -0.40 V
The cell potential at standard conditions (E°cell) can be calculated using the Nernst equation:
E°cell = E°(Cd2+/Cd) - E°(Mn2+/Mn)
E°cell = -0.40 V - (-1.18 V)
E°cell = 0.78 V
Next, we will calculate the cell potential when [Cd2+] reaches 0.035 M using the Nernst equation:
Ecell = E°cell - (0.0592/2) * log([Mn2+] / [Cd2+]) = 0.78 V - (0.0592/2) * log(0.090 / 0.035)
Ecell = 0.78 V - (0.0296) * log(2.57)
Ecell = 0.78 V - (0.0296) * 0.409
Ecell = 0.78 V - 0.0126
Ecell = 0.7674 V
Therefore, the cell potential when [Cd2+] reaches 0.035 M is 0.7674 V.