[ixi]

titrate 50.0 mL 0.05 M Fe2+ vs 0.1 M with Ce4+
compute the Ecell against an NHE for the addition of the following amounts of Ce4+ --> 10 mL

could someone tell me if this is correct? Thanks



[Fe2+]= (0.05 mol/L)(50 mL)(1 L/1000 mL) = 0.0025 mols= 2.5 mmol
[Ce4+]=1 mmol

Fe2+ + Ce4+  ---> Fe3+ + Ce3+
2.5        1
1.5+x    x             1-x        1-x
standard potention (E0) for Fe= 0.771
for Ce=1.61

using Nernst equation:
Efe=0.771-0.059log[Fe2+]/[Fe3+]
ECe=1.61=0.059log[Ce3+]/[Ce4+]

they're equal at equilbrium so i set the 2 equations equal to one another and get:
0.839=0.059log{[Ce3+][Fe3+]}/{[Ce4+][Fe2+]}=log Keq
solving for it--
{[Ce3+][Fe3+]}/{[Ce4+][Fe2+]}=1.66*10^14 =Keq

[(1-x)(1-x)]/[(x)(1.5+X)]=1.66*10^14
ignoring the x cause it's so small--
1/1.5x=1.66*10^14
x=4.0*10^-15 mmol

E= 1.61-0.059log(1/(4.0*10^-1))=0.751 V