**The Nernst Equation at 25****°C:**

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{05916}\mathbf{}\mathbf{V}\mathbf{}}{\mathbf{n}}\mathbf{\right)}{\mathbf{logQ}}}$

E_{cell} = cell potential under non-standard conditions

E°_{cell} = standard cell potential

n = number of e^{-} transferred

Q= reaction quotient = [products]/[reactants]

When writing a cell notation, we use the following format – “* as easy as ABC*”

*Get the overall reaction by balancing the number of electrons transferred then adding the oxidation half-reaction and reduction half-reaction.*

Zn^{2+}_{(aq)} + ** 2 e** → Zn

Zn_{(s)} → Zn^{2+}_{(aq)} + ~~2 e~~^{-}

__________________

Zn_{(s)} + Zn^{2+}_{(aq)}→ Zn^{2+}_{(aq)} + Zn_{(s)}

**# of e- transferred = n = 2 e ^{-}**

The voltage generated by the zinc concentration cell described by,

Zn(s)|Zn2+(aq, 0.100 M)||Zn2+(aq, ? M)|Zn(s)

is 23.0 mV at 25°C. Calculate the concentration of the Zn^{2 +} (aq) ion at the cathode.

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