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Q.Write Nernst's equation for the following cell and find out the emf (electro motive force) of the cell at 298 ∘K298\,{}^\circ K. (Ecell0=2.71 VE^0_{cell} = 2.71\,V) Mg(s) ∣ Mg+2(0.001M) ∣∣ Cu+2(0.0001M) ∣ Cu(s)Mg(s)\,|\,Mg^{+2}(0.001M)\,||\,Cu^{+2}(0.0001M)\,|\,Cu(s)

(OR)
Describe the Lead Accumulator Cell with labelled diagram and write the equation of chemical reactions taking place in it.
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2024Subjective· 3mImportance★★★★★
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Figure — The OR alternative asks to 'Describe the Lead Accumulator Cell with labelled diagram' and the answer supplies
Figure — The OR alternative asks to 'Describe the Lead Accumulator Cell with labelled diagram' and the answer supplies

Applying the Nernst equation with n=2n=2 gives Ecell≈2.68 VE_{cell} \approx 2.68\ V.

For the cell Mg(s) ∣ Mg2+(0.001M) ∣∣ Cu2+(0.0001M) ∣ Cu(s)Mg(s)\,|\,Mg^{2+}(0.001M)\,||\,Cu^{2+}(0.0001M)\,|\,Cu(s), the overall cell reaction is:

Mg(s)+Cu2+(aq)→Mg2+(aq)+Cu(s)Mg(s) + Cu^{2+}(aq) \rightarrow Mg^{2+}(aq) + Cu(s)

with n=2n = 2 electrons transferred.

The Nernst equation at 298 K is:

Ecell=Ecell0−0.0591nlog⁡[Mg2+][Cu2+]E_{cell} = E^0_{cell} - \frac{0.0591}{n}\log\frac{[Mg^{2+}]}{[Cu^{2+}]}

Substituting the given values (Ecell0=2.71 VE^0_{cell}=2.71\,V, [Mg2+]=0.001 M[Mg^{2+}]=0.001\,M, [Cu2+]=0.0001 M[Cu^{2+}]=0.0001\,M):

Ecell=2.71−0.05912log⁡(0.0010.0001)=2.71−0.02955×log⁡(10)E_{cell} = 2.71 - \frac{0.0591}{2}\log\left(\frac{0.001}{0.0001}\right) = 2.71 - 0.02955 \times \log(10)

Ecell=2.71−0.02955×1=2.71−0.02955=2.68045 VE_{cell} = 2.71 - 0.02955 \times 1 = 2.71 - 0.02955 = 2.68045\ V

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