Calculate E, E^o and (delta)G^o for the following cell reaction:
Mg(s) + Sn2+(aq) (yields) Mg2+(aq) + Sn(s)
[Mg2+] = 0.045 M, [Sn2+] = 0.035 M
Mg ==> Mg^+2 + 2e Eo = 2.37 = E1
Sn^+2 + 2e ==> Sn Eo = -0.14 = E2
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Mg + Sn^+2 ==> Mg^+2 + Sn Eo = E1+E2 = 2.23 v.
Note that you should look up these values. My tables are 20 years old.
Ecell = Eocell - (0.0592/n)*log Q.
Q is where you substitute the concns given.
Then delta Go = nFEocell
I was taught that E^o of the cell was the cathode minus anode, but when I do this after changing the signs for reversing the reaction, I get the wrong answer. I only get the right answer when I don't change the sign which doesn't seem right.
To calculate E, E^o, and ΔG^o for the given cell reaction, we need to use the Nernst equation and the standard reduction potentials.
Step 1: Write the half-reactions and their standard reduction potentials:
Mg2+(aq) + 2e- → Mg(s) (E^o = -2.37 V)
Sn2+(aq) + 2e- → Sn(s) (E^o = -0.14 V)
Step 2: Determine the overall cell reaction by summing the half-reactions. In this case, the Mg half-reaction's electrons will be multiplied by 1, and the Sn half-reaction's electrons will be multiplied by 2 to balance the electrons:
2Mg(s) + Sn2+(aq) → 2Mg2+(aq) + Sn(s)
Step 3: Use the Nernst equation to calculate E for the cell reaction:
E = E^o - (RT / nF) * ln(Q)
Where:
- E^o is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- n is the number of electrons transferred
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient
Since the cell reaction is at standard conditions (1 atm and 25°C = 298 K), we can simplify the Nernst equation to:
E = E^o - (0.0257 / n) * ln(Q)
Step 4: Calculate Q by using the concentrations of the species involved in the cell reaction:
Q = [Mg2+]^2 / ([Mg(s)]^2 * [Sn2+]^1)
= (0.045^2) / ((1)^2 * 0.035)
= 0.0354
Step 5: Substitute the given values into the simplified Nernst equation to calculate E:
E = -2.37 V - (0.0257 / 2) * ln(0.0354)
E = -2.37 V - (0.01285) * ln(0.0354)
E ≈ -2.37 V - (-0.1335)
E ≈ -2.37 V + 0.1335
E ≈ -2.2365 V
Step 6: Calculate E^o using the equation:
E^o = E + (RT / nF) * ln(Q)
Since R, T, and F values remain the same, E^o can be determined for standard conditions:
E^o ≈ -2.2365 V + (0.0257 / 2) * ln(0.0354)
E^o ≈ -2.2365 V + (0.01285) * ln(0.0354)
E^o ≈ -2.2365 V + (-0.1335)
E^o ≈ -2.3700 V
Step 7: Calculate ΔG^o using the equation:
ΔG^o = -nF * E^o
= -2 * 96,485 C/mol * -2.3700 V
= 452,664 J/mol
Therefore, the values for E, E^o, and ΔG^o for the given cell reaction are approximately -2.2365 V, -2.3700 V, and 452,664 J/mol, respectively.
To calculate E, E^o, and ΔG^o for the given cell reaction, we need to follow these steps:
Step 1: Write the balanced equation for the cell reaction.
The given cell reaction is:
Mg(s) + Sn2+(aq) → Mg2+(aq) + Sn(s)
Step 2: Identify the half-reactions.
From the balanced equation, we can see that the oxidation half-reaction is the conversion of Mg(s) to Mg2+(aq), and the reduction half-reaction is the conversion of Sn2+(aq) to Sn(s).
Oxidation half-reaction: Mg(s) → Mg2+(aq) + 2e^-
Reduction half-reaction: Sn2+(aq) + 2e^- → Sn(s)
Step 3: Look up the standard reduction potentials (E^o) for each half-reaction.
Using a table of standard reduction potentials, we find that:
E^o(Mg2+/Mg) = -2.37 V
E^o(Sn2+/Sn) = -0.14 V
Step 4: Calculate E^o for the overall reaction.
E^o can be calculated using the equation:
E^o(cell) = E^o(reduction) - E^o(oxidation)
E^o(cell) = E^o(Sn2+/Sn) - E^o(Mg2+/Mg)
E^o(cell) = -0.14 V - (-2.37 V)
E^o(cell) = 2.23 V
Step 5: Calculate E for the overall reaction.
To calculate E, we need to consider the concentrations of the species involved and use the Nernst equation:
E = E^o - (0.0592 V/n) * log(Q)
where Q is the reaction quotient and n is the number of electrons transferred.
In this case, n = 2 as seen in the balanced equation. The reaction quotient (Q) can be calculated using the concentrations of the species involved:
Q = [Mg2+] / [Sn2+]
Q = 0.045 M / 0.035 M
Q = 1.286
Now, we substitute these values into the Nernst equation:
E = 2.23 V - (0.0592 V/2) * log(1.286)
E = 2.23 V - (0.0296) * log(1.286)
E ≈ 2.23 V - (0.0296) * 0.108
E ≈ 2.23 V - 0.00319
E ≈ 2.22681 V
E ≈ 2.23 V (rounded to two decimal places)
Step 6: Calculate ΔG^o for the overall reaction.
ΔG^o can be calculated using the equation:
ΔG^o = -n * F * E^o
where n is the number of electrons transferred and F is the Faraday constant.
In this case, n = 2, and F = 96500 C/mol.
ΔG^o = -(2) * (96500 C/mol) * (2.23 V)
ΔG^o ≈ -4.53 x 10^5 J/mol
ΔG^o ≈ -453 kJ/mol (rounded to three significant figures)
Therefore, the calculated values for the cell reaction are:
E = 2.23 V
E^o = 2.23 V
ΔG^o = -453 kJ/mol