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`

Emf of the cell `Ni| Ni^(2+) ( 0.1 M) | Au^(3+) (1.0M)` Au will be `E_

1.	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» 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`.

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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`

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