Calculate emf of the following cell at 25 °C:…

Calculate emf of the following cell at 25 °C:

Ans.

1. Sn/Sn2+ (0.001 M) and H+/H2(g) (1 bar) | Pt(s)

The overall cell reaction is:

Sn→Sn2++2e−Sn→Sn2++2e−

From the given data:

• �∘(Sn2+/Sn)=−0.14E∘(Sn2+/Sn)=−0.14 V

• �∘(H+/H2)=0.00E∘(H+/H2​)=0.00 V

• $\text{pH}=-\log (0.01)=2$

The reaction quotient ($Q$) for this cell is:

�=[H+]�H2=0.011=0.01Q=PH2​​[H+]​=10.01​=0.01

Now, let’s calculate the emf ($E$) of the cell using the Nernst equation:

�=�∘−����ln⁡(�)E=E∘−nFRT​ln(Q)

Since the cell reaction involves the transfer of 2 electrons ($n=2$), we have:

�=−0.14−(8.314×298)(2×96485)×ln⁡(0.01)E=−0.14−(2×96485)(8.314×298)​×ln(0.01)

Now, let’s compute the value:

�=−0.14−(2475.972)(192970)×ln⁡(0.01)E=−0.14−(192970)(2475.972)​×ln(0.01) �=−0.14−2475.972192970×(−4.605)E=−0.14−1929702475.972​×(−4.605) $E=-0.14+0.005914$ $E=-0.134086\text{ V}$

Therefore, the emf of the given cell at 25°C is approximately -0.134 V.