Calculate the reduction potential of the following solutions:
a- 1 M CuSO4
b- 1 M FeSO4
c- 0.1 M CuSO4
I'm now the same answer as you. :( This is how I did it can you please check.
a- Cu(s)---> Cu(aq) + 2e = -0.34 V
SO4^-2(aq) + 4H(aq) + 2e --> H2SO3(aq) + H2O(l)= 0.20V
Cu(s) + SO4^-2(aq) + 4H(aq) ---> Cu^+2(aq) + H2SO3(aq) + H2O(l)
Ecell= Ecat -Eanode = 0.20-(-0.34)=0.54V
and I did the same for the other two and my answers were:
b- -0.25V
c- 0.451V (but for this one I used the equation : Ecell= Ecell - (0.0592V/n)logQ
I'm really trying but I don't if I'm doing it right.
I'm unsure of what you are doing. If you want the potential of 1M CuSO4, I think that is 0.34v (since 1M is standard that is the number that comes from the table). Same thing for FeSO4 since that is 1 M. It appears to me that you are trying to make a voltaic cell out of just one metal. The SO4^2- is a spectator ion. The reduction potential for 0.1M CuSO4 is
E = Eocell - (0.0592/n)log(reduced)/(oxidized)
E = 0.34 - (0.0592/2)log(Cu/Cu^2+)
E = 0.34 - (0.0296)log(1/0.1)
E = 0.34 - (0.0296)(1)
E = 0.34 - 0.0296 = 0.31 (I'm remembering that I wrote 0.34+0.0296 = 0.37 before. Check this carefully. If I did that it's because I multiplied by -1 and that is wrong.
You're on the right track! Let's go through each solution to calculate the reduction potential.
a) 1 M CuSO4:
First, you correctly wrote down the half-reaction:
Cu(s) -> Cu2+(aq) + 2e- (reduction half-reaction)
Next, you need to look up the standard reduction potential for the Cu2+ to Cu(s) half-reaction. The standard reduction potential (E°) for this reaction is -0.34 V.
b) 1 M FeSO4:
Similarly, you need to write down the half-reaction:
Fe2+(aq) -> Fe(s) + 2e- (reduction half-reaction)
The standard reduction potential for the Fe2+ to Fe(s) half-reaction is -0.44 V. However, since you are working with FeSO4, you need to take into account that the Fe2+ ion will oxidize to Fe3+ in solution. In this case, you also need to consider the additional half-reaction:
Fe3+(aq) + e- -> Fe2+(aq) (oxidation half-reaction)
The standard reduction potential for the Fe3+ to Fe2+ half-reaction is +0.77 V.
To calculate the overall reduction potential, you need to subtract the oxidation potential from the reduction potential:
Reduction potential = Reduction potential of reduction half-reaction - Reduction potential of oxidation half-reaction
Reduction potential = (-0.44 V) - (+0.77 V) = -1.21 V
c) 0.1 M CuSO4:
For this solution, you correctly identified that you can use the Nernst equation to calculate the reduction potential since the concentration is not 1 M. The Nernst equation is:
Ecell = E°cell - (0.0592 V/n)log(Q)
First, let's write down the half-reaction:
Cu(s) -> Cu2+(aq) + 2e- (reduction half-reaction)
Now we need to calculate the value of Q, which represents the reaction quotient. In this case, Q is calculated using the concentrations of the species involved in the reaction. Since you are given a 0.1 M CuSO4 solution, the concentration of Cu2+ is also 0.1 M.
Q = [Cu2+]/[Cu(s)]
Q = 0.1/1 = 0.1
Now we can substitute values into the Nernst equation:
Ecell = E°cell - (0.0592 V/n)log(Q)
Ecell = -0.34 V - (0.0592 V/2)log(0.1)
Ecell = -0.34 V - (0.0296 V)log(0.1)
Ecell ≈ -0.34 V + (0.0296 V) × 1
Ecell ≈ -0.34 V + 0.0296 V
Ecell ≈ -0.27 V
Therefore, the reduction potential for a 0.1 M CuSO4 solution is approximately -0.27 V.
Remember to always carefully check the values coming from reliable sources for the half-reaction potentials.