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The standard emf of the cell, Ni|Ni^(2+)...

The standard 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]

A

`-0.25+0.80=0.55`Volt

B

`-0.25-(+0.80)=-1.05`volt

C

`0+0.80-(-0.25)=+1.05` volt

D

`-0.80-(-0.25)=-0.55`volt

Text Solution

Verified by Experts

The correct Answer is:
C
`E_(cell)=E_(Ni//Ni^(2+))^(@)+E_(Ag^(+))//Ag)^(@)`
`=0.25+0.80=1.05"volt"`
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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 .

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