Use the Nernst equation and data from Appendix D to calculate Ecell for each of the following cells.
Enter your answer with 3 significant digits.
Enter scientific notation as 1.23E4.
(a) Mn(s) | Mn2+(0.11) M || Cr3+(0.24 M), Cr2+(0.11 M) | Pt(s)
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(b) Mg(s) | Mg2+(0.070) M || [Al(OH)4]–(0.30 M), OH–(0.040 M) | Al(s)
To calculate Ecell for each of the given cells, we will use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential (available in Appendix D)
- n is the number of moles of electrons transferred in the balanced equation
- Q is the reaction quotient, which can be calculated using the concentrations of the species involved in the half-cells.
Let's calculate Ecell for each of the given cells:
(a) Mn(s) | Mn2+(0.11) M || Cr3+(0.24 M), Cr2+(0.11 M) | Pt(s)
For this cell, we have the following half-reactions:
Mn2+(aq) + 2e- -> Mn(s) (Reduction half-reaction)
Cr3+(aq) + 3e- -> Cr2+(aq) (Oxidation half-reaction)
Looking up the standard cell potentials for these half-reactions in Appendix D, we find:
E°red = -1.18 V for Mn2+(aq) + 2e- -> Mn(s)
E°ox = -0.74 V for Cr3+(aq) + 3e- -> Cr2+(aq)
The overall cell potential can be calculated as:
E°cell = E°red - E°ox = (-1.18 V) - (-0.74 V) = -0.44 V
Now, let's calculate the reaction quotient (Q):
Q = [Cr2+][Mn2+]/[Cr3+]
Q = (0.11 M * 0.11 M) / (0.24 M) = 0.0505
Using the Nernst equation, we can determine Ecell:
Ecell = -0.44 V - (0.0592/5) * log(0.0505)
Ecell = -0.44 V - 0.01184 * log(0.0505)
Calculating this, we find that Ecell = -0.58 V.
(b) Mg(s) | Mg2+(0.070) M || [Al(OH)4]–(0.30 M), OH–(0.040 M) | Al(s)
For this cell, we have the following half-reactions:
Mg2+(aq) + 2e- -> Mg(s) (Reduction half-reaction)
Al(s) + 3OH-(aq) -> [Al(OH)4]-(aq) (Oxidation half-reaction)
Looking up the standard cell potentials for these half-reactions in Appendix D, we find:
E°red = -2.37 V for Mg2+(aq) + 2e- -> Mg(s)
E°ox = 0.83 V for Al(s) + 3OH-(aq) -> [Al(OH)4]-(aq)
The overall cell potential can be calculated as:
E°cell = E°red - E°ox = (-2.37 V) - (0.83 V) = -3.20 V
Now, let's calculate the reaction quotient (Q):
Q = [Mg2+]/[OH-]^3
Q = 0.070 M / (0.040 M)^3 = 43.75
Using the Nernst equation, we can determine Ecell:
Ecell = -3.20 V - (0.0592/2) * log(43.75)
Ecell = -3.20 V - 0.0296 * log(43.75)
Calculating this, we find that Ecell = -3.77 V.
Therefore, to summarize the calculations:
(a) Ecell = -0.58 V
(b) Ecell = -3.77 V
a) Mn(s) | Mn2+(0.11) M || Cr3+(0.24 M), Cr2+(0.11 M) | Pt(s)
1. Calculate4 the Mn^2+ + e ==> Mn couple.
E = Eo - 0.0592/n)*log(red/(ox)
E = Eo-(0.0592/n)*log[Mn(s)]/(Mn^2+)
E = Eo = (0.0592/2)*log[(1)/(0.11)]
You will need ot look up Eo for this REDUCTION, calculate E for the half cell, then change the sign since this is an oxidation.
Do the same for the Cr^3+ + e ==> Cr^2+
Then EMn as oxidation + ECr as redn = Ecell.