Question

Emf of the cell
`Ni| Ni^(2+) ( 0.1 M) | Au^(3+) (1.0M)` Au will be
`E_(Ni//Ni(2+))^@ = 0.5=25. E_(Au//Au^(3+))^@ = 1.5 V`.
A. ` 1. 75 V`
B. ` +1. 7795 V`
C. ` = 0 . 775 V`
D. `- 1.7795 V`

Answer 1

Correct Answer - B
Cell reaction: `3Ni + 2 A u^(+3) rarr 3 Ni^(+2) = 2 Au`
`E_("cell") = E_("cell")^@ - (0.0591)/6 log. ([Ni^(+2)]^2)/([Au^(+3)]^2)`
` =(0.25+ 15) - (0.0591)/6 log. ((0.1)^3)/((1)^2)`
` = 1.75 - (0.0591) /2 log.(.1)`
`=1. 75 + 0. 295 = +1,7795 V`.