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Mg + Cu^(2+) to Mg^(2+) +Cu , What is E(...

`Mg + Cu^(2+) to Mg^(2+) +Cu` , What is `E_("cell")^(0)` ?
`E_(Mg^(2+)//Mg)^(0)=-2.36 V, E_(Cu^(2+)//Cu)^(0)=0.34 V`

A

`2.7 V`

B

2.7 V

C

`-2.02 V`

D

`2.02 V`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(cell)^(0)=E_(cathode)^(0)-E_(anode)^(0) =0.34-(-2.36)=2.7 V`
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