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The emf of a galvanic cell with electro...

The emf of a galvanic cell with electrode potential of Zn=+0.76 V and that of Cu =-0.34 V is

A

`-1.1 V`

B

`+1.1 v`

C

`+0.34 V`

D

`+0.76 v`

Text Solution

Verified by Experts

The correct Answer is:
A
Given : `Zn^(2+) // Znb =+0.76 V and Cu^(2+)//Cu =- 0.34 V`
`therefore` zn is cathode (`because` of higher reduction potential ) and Cu is znode
`E_(cell) =E_(c )-E_(A)=0.76 -(-0.34)=0.76+0.34 =1.10 V`
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