Check me out on this. The differing systems can be confusing.
The SCE placed on the left SHOULD mean it is the anode and the reaction occurring there is oxidation.
Look up your value of SCE, write it as an oxidation and the appropriate sign. I THINK it is -.241 if written as an oxidation.
For the right hand compartment of the cell, we have the reduction portion which is
E = Eo-(0.0592/n)*log[(Ag)/(Ag^+)]
E = 0.799 -(0.0592/1)log[(1)/(0.02)]
Check out reduction potential for Ag. It may be different from 0.799 in your text. The 0.02 comes from this.
15 mL x 0.05 = 0.75 mmoles Ag^+ initially.
10 mL x 0.025 = 0.25 mmoles Br^- added.
0.75-0.25 = 0.5 mmoles Ag^+ remaining in a volume of 25 mL which is 0.02 M.
Calculate E which is E of the Ag^+ cell.
Then Ecell = Eoxidization half + Eredn half = Ecalomel + EAg
Make sure the calomel is written as an oxidation and the sign is appropriate for the voltage.
Hi
Can someone please help me undersand how to get the answer to this question
A 15.0 mL solution of 0.050 mol/L AgNO3 was titrated with 0.0250 mol/L NaBr in the cell
SCE || titrated solution | Ag(s)
Find the cell voltage for 10.0 mL of titrant added.
(a) -0.0312 V
(b) 0.194 V
(c) 0.457 V
(d) 0.698 V
(e) None of these is correct
1 answer
0 answers