If the Pb+2 concentration is maintained at 1.0M what is the [Cu2+], when the cell potential drops to zero? Ecell0=0.473V; Pb(s)/Pb2+//Cu2+/Cu(s

Ecell= Ecello- 0.05912log [Pb2+][Cu2+] 0=0.473-0.05912log 1.0[Cu2+] log 1[Cu2+]=-0.473-0.02995=16 - log [Cu2+]=16 [Cu2+]=1 X 1016M

If the Pb+2 concentration is maintained at 1.0M what is the [Cu2+], when the cell potential drops to zero? Ecell0=0.473V; Pb(s)/Pb2+//Cu2+/Cu(s

Ecell= Ecello- 0.05912log [Pb2+][Cu2+] 0=0.473-0.05912log 1.0[Cu2+] log 1[Cu2+]=-0.473-0.02995=16 - log [Cu2+]=16 [Cu2+]=1 X 1016M

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If the Pb⁺²concentration is maintained at 1.0M what is the [Cu²⁺], when the cell potential drops to zero? $E_{cell}^{0} = 0.473 V$ ; ${Pb}_{(s)} / {Pb}^{2 +} / / {Cu}^{2 +} / {Cu}_{(s}$

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$1 \times 10^{- 16} M$

$1 \times 10^{15} M$

$1 \times 10^{- 14} M$

$1 \times 10^{14} M$

answer is A.

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Detailed Solution

$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log \frac{[{Pb}^{2 +}]}{[{Cu}^{2 +}]} 0 = 0.473 - \frac{0.0591}{2} \log \frac{1.0}{[{Cu}^{2 +}]} \log \frac{1}{[{Cu}^{2 +}]} = \frac{- 0.473}{- 0.02995} = 16 - \log [{Cu}^{2 +}] = 16 [{Cu}^{2 +}] = 1 X 10^{16} M$