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The emf for the cell Ni|Ni^(2+) (1.0M)|...

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

A

`-0.25+1.50=1.25 V`

B

`-0.25-(+1.50)=-1.75 V)`

C

`1.50-(-(0.25)=1.75 V`

D

`1.50(X3)+(-0.25)X 2=4.0V`

Text Solution

Verified by Experts

The correct Answer is:
C
`E^(0) Ni^(2+)//Ni=-0.25V "and" E^(0) Au^(3+)//Au =+1.5V`
`therefore E_(cell)^(0)=underset(Au)E_(R)^(0)-underset(Ni)E_(R)^(0)=1.50-(-0.25)=1.75`
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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

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]

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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")^@ .

The cell potential for the cell Ni|Ni^(2+)(1M)||Cu^(2+)(1M)|Cu is 0.59 V and E(Cu^(2+)|Cu) = +0.34 V . What is the standard electrode potential of Ni^(2+)|Ni electrode?

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