Be sure to answer all parts.
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 [Mn2+ ] when E
cell
reaches 0.055 V?
To find the concentration of Mn2+ when the cell potential reaches 0.055 V, we first need to calculate the initial standard cell potential (E°cell) of the cell. The standard cell potential can be calculated using the Nernst Equation:
E°cell = E°cathode - E°anode
Given that the standard reduction potentials are E°Mn3+/Mn2+ = -1.18 V and E°Cd2+/Cd = -0.40 V, the standard cell potential can be calculated as follows:
E°cell = E°cathode - E°anode
E°cell = 0.40 V - (-1.18 V)
E°cell = 1.58 V
Next, we can use the Nernst Equation to find the concentration of Mn2+ when the cell potential reaches 0.055 V:
Ecell = E°cell - (0.0592/n)*log(Q)
0.055 V = 1.58 V - (0.0592/2)*log([Mn2+]/[Cd2+])
0.055 V = 1.58 V - 0.0296*log([Mn2+]/0.060 M)
0.5296 = log([Mn2+]/0.060 M)
Antilog(0.5296) = [Mn2+]/0.060 M
3.34 = [Mn2+]/0.060 M
[Mn2+] = 3.34 * 0.060 M
[Mn2+] ≈ 0.2004 M
Therefore, when the cell potential reaches 0.055 V, the concentration of Mn2+ is approximately 0.2004 M.