The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when

Question

3 days ago

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[ Zn2+] = 3.5 M and [Pb2+] = 2.0⋅10−4 M.
Pb2+ (aq) + Zn (s) → Zn2+ (aq) + Pb (s)
A) 0.84
B) 0.76
C) 0.50
D) 0.63
E) 0.39

1 answer:

castortr0y [2.7K] · Answer 1 · 2026-04-01T21:22:23+03:00

Answer: The cell potential for the given reaction stands at 0.50 V

Explanation:

The provided cell reaction is:

$Pb^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Pb(s)$

The half-reactions are:

Anode oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Cathode reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

Initially, we need to find the cell potential for this reaction.

Utilizing the Nernst equation:

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}$

where,

F = Faraday's constant = 96500 C

R = gas constant = 8.314 J/mol·K

T = room temperature = $25^oC=273+25=298K$

n = electrons exchanged in oxidation-reduction = 2

$E^o_{cell}$ = standard electrode potential for the cell = +0.63 V

$E_{cell}$ = cell potential for the reaction =?

$[Zn^{2+}]$ = concentration of Zn²⁺ = 3.5 M

$[Pb^{2+}]$ = $2.0\times 10^{-4}M$

Now substituting all known values into the equation, we arrive at:

$E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}$

$E_{cell}=0.50V$

So, the resulting cell potential for this reaction is 0.50 V