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).
For given cell Anode reaction:
Overall cell reaction:
Here, n = 2,
The Nernst equation for Ecell and 298 k can be written as:
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