From the following reduction potentials, determine the reduction potential of SN^4+ + 4e^- --> Sn^o
SN^2+ + 2e^- --> Sn^o | -0.135V
SN^4+ + 2e^- --> Sn^2+ | -0.15V
Step-by-step please.
To Scott
Please take another look at your calculation in line with this link. I believe the answer is about 0.14. When calculating a half cell potential from two other half cell potentials one cannot simply add the two. This link talks about the why.
https://chemistry.stackexchange.com/questions/9454/deriving-a-reduction-potential-from-two-other-reduction-potentials
I use a short cut that cuts out some of the intermediate steps.
(n1E1 + n2E2)/(n1+n2)
oops ... thanks for the tip
To determine the reduction potential of the reaction SN^4+ + 4e^- --> Sn^o, we can use the concept of reduction potentials and the Nernst equation.
1. Start by writing down the half-reactions for the given reduction potentials:
a. SN^2+ + 2e^- --> Sn^o | -0.135V
b. SN^4+ + 2e^- --> Sn^2+ | -0.15V
2. Notice that we need to balance the half-reaction for the desired reaction. The given half-reactions do not match the desired reaction because they have a different number of electrons involved. So, multiply both half-reactions accordingly to ensure that the number of electrons cancels out:
a. Multiply reaction (a) by 2: 2SN^2+ + 4e^- --> 2Sn^o
b. Multiply reaction (b) by 2: 2SN^4+ + 4e^- --> 2Sn^2+
3. Now we can write the overall balanced equation for the desired reaction by combining the modified half-reactions:
2SN^4+ + 4e^- --> 2Sn^2+ + 4e^- --> 2Sn^o
4. The reduction potential of the overall balanced equation is equal to the sum of the reduction potentials of the individual half-reactions. In this case, we need to add the reduction potentials of reactions (a) and (b) obtained from the given reduction potentials:
Reduction potential of reaction (a): -0.135V
Reduction potential of reaction (b): -0.15V
Reduction potential of the desired reaction:
= Reduction potential of reaction (a) + Reduction potential of reaction (b)
= -0.135V + (-0.15V)
= -0.285V
Therefore, the reduction potential of the reaction SN^4+ + 4e^- --> Sn^o is -0.285V.