A voltaic cell employs the following redox reaction:
Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq)
The half-reactions are as follows:
Sn2+(aq)+2e−→Sn(s) E∘=−0.14
Mn2+(aq)+2e−→Mn(s) E∘=−1.18
Calculate the cell potential at 25 ∘C under each set of conditions.
[Sn2+]= 1.73×10−2 M and [Mn2+]= 1.75 M
[Sn2+]= 1.75 M and [Mn2+]= 1.73×10−2 M
Ered Sn = Eored - (0.0592/2)*log(Sn/Sn^2+)
Ered Sn = -0.14 - (0.0592/2)*log (1/1.73E-2 M). You solve it.
Ered Mn = -1.18 - (0.0591/2)*log (1/1.75 M). You solve it.
Then Ecell = Eored + Eoox.
Eored is Sn and that goes in as calculated above.
Eoox is Mn and you turn the reaction above around AND change sign to +
Add the two to get the cell reaction and the Ecell.
Post your work if you get stuck.
Personally, I think it is easier to do it as follows and I recommend it to you.
Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq)
Sn^2+(aq) + 2e => Sn(s) ..........Eored = -0.14 v
Mn(s) ==> Mn^2+ + 2e ..............Eoox = +1.18 v
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Sn^2+(aq) + Mn(s) ==> Sn(s) + Mn^2+(aq) Eocell = 1.04 v
That is the equation you started with except now is standard Eo cell potential.
Now use the Nernst equation to correct for the fact it isn't standard concentrations.
Ecell = Eo cell - (0.0592/2)*log (Qrxn) and plug in as follows;
For Eo cell you have that above as +1.04 v. For Qrxn it is (products)/(reactants)
Qrxn = [Sn(s)][Mn^2+]/[Sn^2+][Mn(s)] = (1)(1.75)/(1.73E-2)(1)]
You see I plugged in the first set of conditions. Where did the 1 come from? For the pure element we plug in 1 for any pure element which is standard. Hope this helps.
To calculate the cell potential at 25∘C under each set of conditions, we can use the Nernst equation:
E = E° - (0.0592 / n) * log(Q)
where:
E is the cell potential at 25∘C,
E° is the standard cell potential,
n is the number of moles of electrons transferred in the balanced redox reaction,
Q is the reaction quotient, and
0.0592 is the value of RT/F at 25∘C, where R is the gas constant and F is Faraday's constant.
Let's calculate the cell potential for each set of conditions:
Set of conditions 1:
[Sn2+] = 1.73×10^(-2) M and [Mn2+] = 1.75 M
In this case, we have:
Q = [Sn2+]/[Mn2+] = (1.73×10^(-2)) / (1.75) = 9.894×10^(-3)
Using the given standard cell potentials:
E°(Sn2+/Sn) = -0.14 V
E°(Mn2+/Mn) = -1.18 V
The number of moles of electrons transferred in the balanced reaction is 2.
Substituting into the Nernst equation:
E = (-0.14) - (0.0592 / 2) * log(9.894×10^(-3))
Calculating this expression will give us the value of E for set of conditions 1.
Set of conditions 2:
[Sn2+] = 1.75 M and [Mn2+] = 1.73×10^(-2) M
In this case, we have:
Q = [Sn2+]/[Mn2+] = (1.75) / (1.73×10^(-2)) = 101.16
Again using the given standard cell potentials and number of moles of electrons transferred, we can substitute into the Nernst equation to calculate the value of E for set of conditions 2.
Remember to use natural logarithms (ln) in the Nernst equation, as the log function usually refers to base 10 logarithms.
This process will give us the cell potential under each set of conditions.