Question

Il coll velly your subscription Sign In Problem 18 (20 pts) A galvanic cell consists of an electrode of Pb in a 0.6 M solution of PbSO4 and an electrode of Sn in a solution of 0.3 M SnSO4 at 25°C. What is the emf of the cell? emf = 1 8 - D. Focus Display Settings

Answer 1

Cell is galvanic cell means reaction in the cell is spontaneous.

We know the significance of electrode potential that, element having more positive value of reduction potential acts as a cathode & element having less positive value of reduction potential acts as a anode.

If we compare reduction potentials of Pb & Sn, we found that Pb will act as a cathode & Sn will acts as a anode.

Consider a half reactions.

Reaction at anode : Sn (s) $\rightarrow$ Sn ²⁺ (aq) + 2 e ^-

Reaction at cathode : Pb ²⁺ (aq) + 2 e ^-$\rightarrow$ Pb (s)

Overall reaction : Sn (s) + Pb ²⁺ (aq) $\rightarrow$ Sn ²⁺ (aq) + Pb (s)

Number of electrons transferred in the overall reaction ( n) = 2

The standard emf of the cell is calculated as E ⁰ cell = E ⁰ cathode - E ⁰ anode  

E ⁰ cell = - 0.126 V - ( - 0.141 V ) = 0.015 V

The emf of the cell is calculated by using Nernst equation .

E cell = E ⁰ cell - 2.303 R T / n F log Q

E cell = 0.015  - 2.303 R T / n F log [Sn ²⁺] / [ Pb ²⁺]

At 298 K , 2.303 R T / F = 0.0591.

we have , [Sn ²⁺] = 0.3 M & [ Pb ²⁺] = 0.6 M

$\therefore$ E cell = 0.015 - 0.0591 / 2 log 0.3 / 0.6

E cell = 0.015 - ( - 0.00890 )

E cell = 0.015 + 0.00890 V

E cell = 0.0239 V

ANSWER : Emf of the galvanic cell is 0.0239 V