1.

Calculate the emf of the cell in which the following reaction takes place Ni(s) + 2Ag^(+)(0.002M) to Ni^(2+) (0.160 M)+ 2Ag(s) Given that E_("cell")^(@) = 1.05 V.

Answer»

SOLUTION :Applying Nernst equation to the given CELL reaction:
`E_("cell") = E_("cell")^@ - (0.0591)/x LOG (NI^(2+))/([Ag^+]^2)= 1.05 V - (0.0591)/2 log (0.160)/((0.002)^2)`
`= 1.05 - (0.0591)/2 log (4 xx 10^(4)) = 1.05 - (0.0591)/(2) xx 4.602`
`= 1.05 - 0.14 V = 0.91 V`.


Discussion

No Comment Found

Related InterviewSolutions