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E(ni^(2+)//Ni)^(@)=-0.25 VE(Au^(3+)//Au)...

`E_(ni^(2+)//Ni)^(@)=-0.25 VE_(Au^(3+)//Au)^(@)=1.50 v`
The emf of the voltaic cell
`Ni//Ni^(2+)(1.0 M)||Au^(3+)(1.0 M)|Au` is

A

1.25 V

B

`-1.75 V`

C

1.75 V

D

2.0 V

Text Solution

Verified by Experts

The correct Answer is:
C
EMF of the cell = `1.50 -(-0.25) = 1.75 V`
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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:-

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+) (1.0 M)|Au is :

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If E_(Au^(+)//Au)^(@) is 1.69 V and E_(Au^(3+)//Au)^(@) is 1.40 V, then E_(Au^(+)//Au^(3+))^(@) will be :

Calculate emf of the following cell reaction at 2968 K : Ni(s)//Ni^(2+)(0.01 M) || Cu^(2+)(0.1 M)//Cu(s) [Given E_(Ni^(2+)//Ni)^(@)=-0.25" V ",E_(Cu^(2+)//Cu)^(@)=+0.34 " V " ] Write the overall cell reaction.

MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-ELECTROCHEMISTRY-Exercise 1
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