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
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I've always used a shortcut.
(n1*E1 + n2*E2)/(n1+n2)
To determine the reduction potential of SN^4+ + 4e^- --> Sn^o, you can use the Nernst equation.
The Nernst equation relates the standard reduction potential (E°) to the actual reduction potential (E) at a specific concentration of the species involved in the reaction. The equation is as follows:
E = E° - (0.0592 / n) * log(Q)
Where:
E = actual reduction potential
E° = standard reduction potential
n = number of electrons involved in the reaction
Q = reaction quotient
In this case, the reaction is SN^4+ + 4e^- --> Sn^o, and we need to find the reduction potential (E) given the standard reduction potentials provided.
Since we have the reduction potential for the reaction SN^4+ + 2e^- --> Sn^2+ and the reduction potential for the reaction SN^2+ + 2e^- --> Sn^o, we can use them to calculate the reduction potential of the given reaction.
First, we need to find the reduction potential for the reaction SN^4+ + 2e^- --> Sn^o. Since the given reaction is the sum of two half-reactions, we can add their reduction potentials:
SN^4+ + 2e^- --> Sn^2+ | -0.15V
SN^2+ + 2e^- --> Sn^o | -0.135V
Adding these reduction potentials gives us:
-0.15V + (-0.135V) = -0.285V
Now, we have the reduction potential for the reaction SN^4+ + 2e^- --> Sn^o as -0.285V.
To determine the reduction potential of SN^4+ + 4e^- --> Sn^o, we can use the Nernst equation. Since the number of electrons involved in the reaction is 4 (as given in the reaction equation), we can substitute the values into the equation:
E = E° - (0.0592 / n) * log(Q)
E = -0.285V - (0.0592 / 4) * log(1)
E = -0.285V
Therefore, the reduction potential of the reaction SN^4+ + 4e^- --> Sn^o is -0.285V.