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The standard reduction potentials provide potentials for solutes at 1 M or 1 atm partial pressure. At other conditions the next section gives us some idea as to what we might predict for non-standard conditions.

You'll see the **Nernst equation**:

also written as:

where RT/F at 298.15 K = (8.3145 J/mol·K)(298.15 K)/(96485.34 C/mol) = 0.02569 V and 2.303*log* replaces *ln*. (Note that J·C is equivalent to a Volt.) We can use either of these forms to:

- calculate the voltage of an electrochemical cell under any conditions
- to find the concentration of one of the components in a system

Let's consider the reaction of tin and bromine again when the tin(II) concentration is 0.050 M and the tin(IV) and bromide concentrations = 0.00010 M.

Sn^{2+}_{(aq)} + Br_{2}_{ (l)} Sn^{4+}_{(aq)} + 2 Br^{-}_{(aq)}

Q = [Sn^{4+}][Br^{-}]^{2} / [Sn^{2+}]

(b) The Nernst equation for this reaction is:

0.0591V [Sn^{4+}] [Br^{-}]^{2}E = 0.92 V - -------log( ------------ ) 2 [Sn^{2+}] 0.0591V (0.00010 M)(0.00010 M)^{2}E = 0.92 V - -------log( ----------------------- ) 2 0.050 M 0.0591V E = 0.92 V - -------log(2.0x10^{-11}) 2

E = 0.92 V - (-0.316 V)

E = 1.24 V

Consider two beakers containing silver nitrate solutions of different concentrations, say 0.010 M and 0.50 M. If we connect the two beakers do we have a system at equilibrium?

No, there is a concentration gradient and there is a driving force for diffusion to attain equal concentrations in both beakers. We can set up an electrochemical cell to capture the chemical potential of this non-equilibrium system.

| | | | |---------|_____|---------| | _____ | | 0.01 M | | 0.5 M | | Ag+ | | Ag+ | |_________| |_________|

The two half-reactions are:

Ag_{(s)} Ag^{+}_{(aq)} + e^{-}
Ag^{+}_{(aq)} + e^{-} Ag_{(s)}

Q = [Ag^{+}] / [Ag^{+}] (To determine which concentration goes where, on which side of the half-reactions are the silver ion.)

(b) The Nernst equation for this reaction is:

0.0591V 0.010 M E = 0.0 V - -------log( ------- ) 1 0.50 M E = 0.0 V - (- 0.100 V)

E = 0.10 V

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