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A concentration cell consists of two H2/H half-cells. Half cell A has H2 0.95atm bubbling into 0.10M HCl. Half-cell B has H2 at0.60atm bubbling into2.0M HCl. Which half-cell houses the anode? What is the voltage of the cell?

Short Answer

Expert verified

Anode: Half cell A

Ecell=0.0385V

Step by step solution

01

To find the voltage of the cell

Half cell A:2H+2e-→H2(O.95atm,0.10MHCI)

Half cell B:2H++2e-→H2(0.60atm,2.0MHCI)

Oxidation takes place at anode, it has a negative charge. Therefore, electrode A is negative and electrode B is positive.

In a concentration cell, the anode half-cell contains the more dilute solution, which is the solution in half-cell A.

Hence, we get the following cell reactions:

Anode:H2→2Hϕ+2e-;(0.95atm,0,10MHCI)

Cathode:2H+2e-→H2;(0.60atm,2,0NHCI)

Cell reaction:H+(aq;2.0M)halfcellB$→H+(aq;0.10M)/halfcellA

Ecell=Ecell-0.0592nlogQ∘=0-0.05922log0.01M2.0M=-0.05922×(-2.3)=0.0385V

Hence Anode: Half cell A

Ecell=0.0385V

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