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Chemistry Question on Nernst Equation

The cell potential of the following cell at 25 oC\text{25}{{\,}^{\text{o}}}\text{C} (in volts) is (pt)H2(1 atm) ∣H+(0.01M) ∣∣Cu2+0.1 M ∣Cu\underset{(1\,atm)}{\mathop{(pt){{H}_{2}}}}\,\left| \underset{(0.01M)}{\mathop{{{H}^{+}}}}\, \right|\left| \underset{0.1\,M}{\mathop{C{{u}^{2+}}}}\, \right|Cu (ECu2+∣Cuo=0.337 V)(E_{C{{u}^{2+}}|Cu}^{o}=0.337\,V)

A

0.3080.308

B

0.4270.427

C

−0.308-0.308

D

0.3370.337

Answer

0.4270.427

Explanation

Solution

(Pt)H2(1 atm) ∣H(0.01 M) +∣∣Cu2+(0.1 M) ∣Cu\underset{(1\,atm)}{\mathop{(Pt){{H}_{2}}}}\,\left| {{\underset{(0.01\,M)}{\mathop{H}}\,}^{+}} \right|\left| \underset{(0.1\,M)}{\mathop{C{{u}^{2+}}}}\, \right|Cu For the above cell, the cell reaction is H2+Cu2+→2H++Cu{{H}_{2}}+C{{u}^{2+}}\xrightarrow{{}}2{{H}^{+}}+Cu According to the Nemst equation Ecell=ECu2+/Cuo−0.0591nlog⁡[H+]2[Cu2+]{{E}_{cell}}=E_{C{{u}^{2+}}/Cu}^{o}-\frac{0.0591}{n}\log \frac{{{[{{H}^{+}}]}^{2}}}{[C{{u}^{2+}}]} =0.337−0.05912log⁡(0.01)2(0.1)=0.337-\frac{0.0591}{2}\log \frac{{{(0.01)}^{2}}}{(0.1)} =0.337−0.02955log⁡10−3=0.337-0.02955\log {{10}^{-3}} =0.337+0.08865=0.337+0.08865 =0.42565≈0.427=0.42565\approx 0.427