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 ?

`2.00` V

`1.25 V `

`-1.25 V `

`1.75` V

Text Solution

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The correct Answer is:

Here Ni is anode and Au is cathode
Given `E_(Ni^(2+) // Ni)^(@) = -0.25 V ` and `E_(Au^(3+) // Au)^(@) = + 1.5 V `
`E_(cell)^(@) = E_(C)^(@) - E_(A)^(@) = 1.5 - (-0.25) = 1.5 + 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:-

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 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.

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