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Question 54 of 73

Q.The e.m.f. of the half cell Cu(aq)2+/Cu(s)Cu^{2+}_{(aq)}/Cu_{(s)} containing 0.01 M Cu2+Cu^{2+} solution is +0.301+0.301 V. Calculate the standard e.m.f. of the half cell.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2017Subjective· 5mImportance★★★★★
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Applying the Nernst equation to the given half-cell potential and concentration and solving for E∘E^\circ gives a standard electrode potential of about +0.36+0.36 V for Cu2+/CuCu^{2+}/Cu.

Half-cell reaction (reduction):

Cu2+(aq)+2e−→Cu(s)Cu^{2+}(aq) + 2e^- \rightarrow Cu(s), so n=2n = 2 electrons.

Nernst equation for this half-cell (at 298 K, using 2.303RTF=0.0591\dfrac{2.303RT}{F} = 0.0591 V):

E=E∘−0.0591nlog⁡1[Cu2+]E = E^\circ - \dfrac{0.0591}{n}\log\dfrac{1}{[Cu^{2+}]}

Given data: E=+0.301E = +0.301 V, [Cu2+]=0.01[Cu^{2+}] = 0.01 M =10−2= 10^{-2} M, n=2n = 2.

Substituting:

0.301=E∘−0.05912log⁡110−20.301 = E^\circ - \dfrac{0.0591}{2}\log\dfrac{1}{10^{-2}}

log⁡110−2=log⁡(102)=2\log\dfrac{1}{10^{-2}} = \log(10^{2}) = 2

0.301=E∘−0.05912×20.301 = E^\circ - \dfrac{0.0591}{2}\times 2

0.301=E∘−0.05910.301 = E^\circ - 0.0591

Solving for E∘E^\circ:

E∘=0.301+0.0591=0.3601 VE^\circ = 0.301 + 0.0591 = 0.3601\ \text{V}

Rounding to three significant figures, E∘≈0.36E^\circ \approx 0.36 V.

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