Calculate EMF of the following cells at `25^(@)C`
(i) `Fe|Fe^(2+) (a=0.6 M) || Sn^(2+). (a=0.2M)|Sn`.
`E_(Fe^(2+), Fe)^(0) =-0.44 V and E_(Sn^(2+),Sn)^(0)=0.14V`
(ii) `Pt, H_(2) (3"atm") |HCl |H_(2) ("15 atm"), Pt, E_(H^(+)//1/2 H_(2))^(0)=0V`

Explanation

To calculate the EMF of the given electrochemical cells, we will follow these steps for each part of the question.

### Part (i): Calculation of EMF for the cell `Fe|Fe^(2+) (a=0.6 M) || Sn^(2+). (a=0.2M)|Sn`

1. **Identify the half-reactions and standard electrode potentials:**
- For the oxidation half-reaction at the anode:
\[
\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \quad (E^\circ = +0.44 \, \text{V})
...