Question

For a cell reaction in involving a two electron change, the Stanford e. m. f of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be

विकल्प

• 2.95 × 10^–2

• 10

• 1 × 10^–10

• 2.95 × 10^–10

Answer 1

10

Explanation:

Using the relation,

`E_(cell)^circ = (2.303  RT)/(nF)  log K_c = 0.0591/n  log K_c`

∴ 0.295 V = `0.0591/2  log K_c`

or `log K_c = (2 xx 0.295)/0.0591` = 10

or `K_c = 1 xx 10^10`