Question

Given the data at 25°C

Ag + I^– → Agl + e^–; E° = – 0.152 V

Ag → Ag⁺ + e^–; E° = – 0.800 V

The value of log K_sp for Ag I is :-

Options

• – 8.12

• + 8.162

• – 37.83

• – 16.13

Answer 1

– 16.13

Explanation:

Agl⁺ + e^– Ag + I^– > E° = – 0.192

Ag → Ag⁺ + e^–; E° = – 0.800

Agl →Ag⁺ + I^– ; E° = – 0.952

`E_(cell)^circ = 0.059/1  log  K =  0.059/1  log K_(sp)`

log K_sp = `(E_(cell)^circ)/0.059 = ((-0.952 xx 1))/((0.059))` = – 16.13