Question

10.0 ml solution of 0.050 m agno3was titrated with 0.0250m nabr in the cell s.E.C find the cell voltage for 30.0 ml of titrant

Answer 1

Answer:

A

Explanation:

Answer 2

Answer: The cell voltage for 30.0ml of titrant is -0.039V.

Explanation:

Ksp of AgBr = 5.0 x 10-13

The required equation is,

KBr + AgNO3 -->  AgBr + KNO3

When 30 ml titrant is added

Ve =20 ml

Ag+ +e- --> Ag (s)

Eº =0.799 V

E = Eº -RT/nF ln Q

E= 0.799 -0.05916 log (1/ [Ag+] )

ESCE=0.241 V

30 ml NaBr ;

     Ag+(ionic state)  +  Br- (ionic state)    =  AgBr(s)

S   0.5 mmol           0.75              0

R  -0.5                     -0.5             +0.5

F   0                         0.25/40 =0.625 M

Ag+ =KSP/[Br-] = 5.00 x 10-13 /0.00625 =8.0 x 10-11M

E=E+  - E- = {0.799 -0.05916 log 1/ 8 x 10-11} -0.241

E = -0.039V

Hence, the cell voltage for 30.0 ml of titrant is -0.039V.

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