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Ni|Ni^(2+)(1.0M)||Au^(3+)(1.0M)| Au (whe...

`Ni|Ni^(2+)(1.0M)||Au^(3+)(1.0M)|` Au (where `E^(@)` for ` Ni^(2+)//Niis -0.25and V and E^(@)` for `Au^(3+)//Au` is `(0.150V).` What is the emf of the cell ?

A

`+0.4V`

B

`-1.75 V`

C

`+1.25V`

D

`+1.75 V`

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The correct Answer is:
A
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The emf of the cell, Ni|Ni^(2+)(1.0M)||Ag^(+)(1.0M)|Ag [E^(@) for Ni^(2+)//Ni =- 0.25 volt, E^(@) for Ag^(+)//Ag = 0.80 volt] is given by : [E^(@) for Ag^(+)//Ag = 0.80 volt]

The cell emf for the cell Ni(s)abs(Ni^(2+)(1.0M))Au^(3+)(1.0M)| Au(s) (E^(o) for Ni^(2+) abs(Ni=-0.25 V,E^(o)" for " Au^(3+))Au=1.50V) is

These question consist of two statements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses : Ni//Ni^(2+) (1.0 M) || Au^(3+) (1.0 M) | Au , for this cell emf is 1. 75 V if E_(Au^(3+)//Au)^@ =1.50 and E_(Ni^(3+)//Ni)^2 =0.25 V . Emf of the cell =E_("cathode")^@- E_("anode")^@ .

E^(@)(Ni^(2+)//Ni) =- 0.25 volt, E^(@)(Au^(3+)//Au) = 1.50 volt. The emf of the voltaic cell Ni|Ni^(2+) (1.0M)||Au^(3+)(1.0M)|Au is:-

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 .

MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-PRACTICE SET 21-PAPER 1 (PHYSICS & CHEMISTRY)
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