**Modified Nernst equation**:

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{-}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}}{\mathbf{n}}\mathbf{\right)}{\mathbf{log}}\frac{\left[\mathrm{anode}\right]}{\left[\mathrm{cathode}\right]}}$

• **↓ E°** → oxidation → anode

• **↑ E° **→ reduction → cathode

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn^{2+}(aq) + 2 e^{-} and the cathode half-reaction is Sn^{2+}(aq) + 2 e^{–} → Sn(s). What is the concentration of Sn^{2+} if Zn^{2+} is 2.5 X 10^{-3} M and the cell emf is 0.660 V? Use the reduction potentials in Appendix E that are reported to three significant figures.

(a) 3.3 X 10^{-2} M

(b) 1.9 X 10^{-4} M

(c) 9.0 X 10^{-3} M

(d) 6.9 X 10^{-4} M

(e) 7.6 X 10^{-3} M

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