Question

The emf of the galvanic cell having representation Pb/Pb2+(0.002M)//Pb2+(0.02M)/Pb is approximately? Given E°pb2+/pb = -0.13V.

Answer 1

The EMF of the cell is +0.0295 V

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs at an anodic electrode.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs at a cathodic electrode.

Oxidation half-reaction (anode):     $Pb(s)\rightarrow Pb^{2+}(0.002M,aq)+2e-$

Reduction half-reaction (cathode): $Pb^{2+}(0.02M,aq) + 2e^-\rightarrow→ Pb(s)$

Overall reaction:  $Pb^{2+}(0.02M,aq)\rightarrow Pb^{2+}(0.002M,aq)$

The standard EMF of the cell will be equal to 1 as both the half reactions are equivalent

Nernst equation follows:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log (\frac{\text{[Products]}}{\text{[Reactants]}})$

where,

$E_{cell}$ = electrode potential of the cell

n = number of electrons exchanged = 2

[Products] = concentration of the products = 0.002 M

[Reactants] = concentration of the reactants = 0.02 M

Putting values in above equation, we get:

$E_{cell}=0-\frac{0.059}{2}\log (\frac{0.002}{0.02})\\\\E_{cell}=+0.0295V$

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Answer 2

Answer:

Explanation:see the sol