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The emf of the cell Ni|Ni^(2)(1.0 M)||A...

The emf of the cell `Ni|Ni^(2)(1.0 M)||Ag^(+)(1.0 M)|Ag`
`E^(@)` for `Ni^(2+)//Ni=-0.25 V, E^(@)` for `Ag^(+)//Ag^(+)//Ag=0.80 V` is given by

A

`+0.55 V`

B

`-1.05 V`

C

`+1.05 V`

D

`-0.55 V`

Text Solution

Verified by Experts

The correct Answer is:
A
CMF of the cell =0.80 -(-0.25)
=+1.05 V
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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

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 ?

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 .

Calculate the e.m.f. of the cell, Mg//Mg^(2+)(0.1 M)||Ag^(+)(1.0xx10^(-3)M)//Ag The values of E_(Mg^(2+)//Mg)^(@) and E_(Ag^(+)//Ag)^(@) are -2.37 " V" and +0.80 " V " respectively.

MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-ELECTROCHEMISTRY-Exercise 2
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