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Electode potential of Zn^(2+) //Zn is ...

Electode potential of ` Zn^(2+) //Zn` is ` -0 . 7 V` and that of `Cu^(2+) //Cu si = 0. 3 4 V`. The EMF of the cell consttructed between these two elctrodes is .

A

1.10 V

B

0.42 V

C

`-1.1 V`

D

`-0.42 V`

Text Solution

Verified by Experts

The correct Answer is:
A
`E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@) = 0.34 - (-0.76) = 1.10 V`
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