Question

Calculate the emf of the following cell:

\[\ce{Ni_{(s)} + 2Ag+(0.01 M) -> Ni^{2+} (0.1 M) + 2Ag_{(s)}}\]

Given that `E_(cell)^0` = 1.05 V, log 10 = 1

Answer 1

Using the Nernst equation:

`E_(cell) = E_(cell)^0 - 0.0592/n log  [["Products"]]/[["Reactants"]]`

= `1.05 - 0.0592/2 log  [["Ni"^(2+)]]/["Ag"^+]^2`

= `1.05 - 0.0291 log  0.1/(0.01)^2`

= 1.05 − 0.0291 log 10³   ...[∴ log 10 = 1]

= 1.05 − 0.0291 × 3

= 1.05 − 0.0873

E_cell = 0.962 V