The emf of a cell corresponding to the reaction
`Zn +2H^(+)(aq) rarr Zn^(2+) (0.1M) +H_(2)(g) 1` atm is `0.28` volt at `25^(@)C`. Calculate the `pH` of the solution at the hydrogen electrode.
`E_(Zn^(2+)//Zn)^(@) =- 0.76` volt and `E_(H^(+)//H_(2))^(@) = 0`

Explanation

To solve the problem, we will use the Nernst equation to relate the cell potential (emf) to the concentrations of the reactants and products involved in the electrochemical reaction.

### Step-by-Step Solution:

1. **Identify the Cell Reaction**:
The cell reaction is given as:
\[
\text{Zn} + 2\text{H}^+ \rightarrow \text{Zn}^{2+} (0.1 \, \text{M}) + \text{H}_2 (1 \, \text{atm})
...