The Nernst Equation Video Lessons

Concept:

# Problem: Consider a voltaic cell where the anode half-reaction is Zn(s)  →  Zn2+(aq) + 2 e- and the cathode half-reaction is Sn2+(aq) + 2 e–  →  Sn(s). What is the concentration of Sn2+ if Zn2+ 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 Marray{ { m Zn}^{+2}({ m aq})~+~2~{ m e}^-~&amp;amp; ightarrow &amp;amp;~ m Zn(s)~&amp;amp;~E^{circ}_{ m red}~&amp;amp;=&amp;amp;~-0.76~{ m V} cr { m Sn}^{2+}({ m aq})~+~2~{ m e}^–~&amp;amp; ightarrow &amp;amp;~ m Sn(s)~&amp;amp;~E^{circ}_{ m red}~&amp;amp;=&amp;amp;~-0.136~{ m V}}

###### FREE Expert Solution

Modified Nernst equation:

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

↓ E° → oxidation → anode
↑ E° → reduction → cathode

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###### Problem Details

Consider a voltaic cell where the anode half-reaction is Zn(s)  →  Zn2+(aq) + 2 e- and the cathode half-reaction is Sn2+(aq) + 2 e  →  Sn(s). What is the concentration of Sn2+ if Zn2+ 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