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Calculate for the Electrochemical Cell Below,
Pb(s)|PbCl2(s)| Cl-(aq,1 \to Pb(s)        ~~~ ~~~~

question 51

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Calculate  Calculate   for the electrochemical cell below, Pb(s) |PbCl<sub>2</sub>(s) | Cl<sup>-</sup>(aq,1.0 M) || Fe<sup>3+</sup>(aq,1.0 M) ,Fe<sup>2+</sup>(aq,1.0 M) | Pt(s)  Given the following reduction half-reactions.  Pb<sup>2+</sup>(aq) + 2 e<sup>-</sup>  \to  Pb(s)  ~~~ ~~~~~~~ ~~~~~~~~~~~~ E<sup> \circ </sup> = -0.126 V PbCl<sub>2</sub>(s) + 2 e<sup>-</sup>  \to  Pb(s) + 2 Cl<sup>-</sup>(aq)  ~~~~~~~~ E<sup> \circ </sup> = -0.267 V Fe<sup>3+</sup>(aq) + e<sup>-</sup>  \to  Fe<sup>2+</sup>(aq)  ~~~~ ~~~~~~~~~~~~~~~~~ E<sup> \circ </sup> = +0.771 V Fe<sup>2+</sup>(aq) + e<sup>-</sup>  \to  Fe(s)  ~~~~~~~~~~ ~~~~~~ ~ ~~~~~~~~~ E<sup> \circ </sup> = -0.44 V A)  -0.504 V B)  -0.062 V C)  +0.504 V D)  +1.038 V E)  +1.604 V for the electrochemical cell below,
Pb(s) |PbCl2(s) | Cl-(aq,1.0 M) || Fe3+(aq,1.0 M) ,Fe2+(aq,1.0 M) | Pt(s)
Given the following reduction half-reactions.

Pb2+(aq) + 2 e- \to Pb(s)        ~~~ ~~~~       ~~~ ~~~~        ~~~~~~~~ E \circ = -0.126 V
PbCl2(s) + 2 e- \to Pb(s) + 2 Cl-(aq)         ~~~~~~~~ E \circ = -0.267 V
Fe3+(aq) + e- \to Fe2+(aq)      ~~~~ ~        ~~~~~~~~        ~~~~~~~~ E \circ = +0.771 V
Fe2+(aq) + e- \to Fe(s)         ~~~~~~~~       ~~ ~~~~~   ~ ~ ~        ~~~~~~~~ E \circ = -0.44 V


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