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Electrochemistry

Question
CBSEENCH12006023
Wired Faculty App

Write the Nernst equation and emf of the following cells at 298 k.
Pt(s) | Br2 (l) | Br– (0.010 M) || H+ (0.030 M) | + H2(g) (1 bar) | Pt(s).

Solution
For given cell Anode reaction:

2Br-(aq) → Br2(l) + 2e-

Cathode reaction:

2H+(aq) + 2e- → H2(g)

Overall cell reaction:
                     2Br-(aq) + 2H+(aq) → Br2(l) + H2(g)

Here,  n = 2,  

E0cell = E0cathode -E0anode = 0- 1.08 V = -1.08 V.

The Nernst equation for Ecell and 298 k can be written as:

Ecell  =E0cell - 0.059nlog Br2+ H2Br-H+2       = 1.08 - 0.0592 log 1[10-2]2 [ 3 × 10-2] 2        =-1.08 - 0.0295 log 110-4×9×10-4        = -1.08 - 0.0295 (log 108 - log 9)         = - 1.08 -  0.0295 (7.0458)          = -1.08- 0.2078 = -1.2878 V.

The negative value of Ecell indicates the cell has been arranged in a reverse way, i.e., hydrogen electrode will act as anode and bromine electrode act as cathode. The cell should be represented as Pt | H2 (1 bar), H+ (0.03 M) || Br (0.01 M) | Br2(l), Pt

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