Question

Calculate the electrode potentials of the following half cells at 298 K.

Ag_(s) | AgNO₃ (0.1 M)

Given: \[\ce{E^{\circ}_{Ag^+/Ag}}\] = +0.80 V, \[\ce{E^{\circ}_{{Co^{2+}/Co}}}\] = −0.28 V

Answer 1

Given: Standard electrode potential (E°) = +0.80 V

Ag⁺ = 0.1 M

T = 298 K

n = 1

The nernst equation is 

E = `E^circ - (RT)/(nF) log Q`

For the given reaction Q = `1/([text{Ag}^+])`

At 298 K, `(RT)/F` = 0.0592 V

So, E = `E^circ - 0.0592/n log_10  1/0.1`

E = `0.80 V - 0.0592/1 log_10  10`

E = 0.80 V − 0.0592 × 1

E = 0.80 V − 0.0592

E = 0.7408 V