Question

Calculate the EMF of the cell
Cr | Cr+3 (0.1M) || Fe+2 (0.01M) I Fe
(Given E^0cr+3/cr = -0.75V, E^0 Fe^+2/ Fe = -0.45 V)​

Answer 1

Answer:

The reaction taking place in the cell can be balanced as:

2Cr + 3Fe+2 -----------> 2Cr+3 + 3Fe

So, n = 6

Now E°cell = E°cathode - E°anode = -0.45-(-0.75) = 0.3V

As concentration of Cr+3 and Fe+2 are not 1M, so EMF can be determined using Nernst equation.

So, Ecell = E°cell - (0.059/6) log {[Cr+3]^2/[Fe+2]^3}

= 0.3 - (0.059/6) log {[0.1]^2/[0.01]^3}

= 0.3 - (0.059/6) log (10^(-4))

= 0.3 + (4×0.059)/6

= (0.9 + 0.118)/3

= 1.018/6 = 0.336V

I hope it's correct.