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 (Electromotive Force) of the given electrochemical cells at 25°C, we will use the Nernst equation. The Nernst equation is given by:

\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

where:
- \(E_{\text{cell}}\) is the cell potential under non-standard conditions.
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