Question

Calculate emf of the following cell Zn(s)/Zn2+(0.1m) l l (0.01m)Ag+/Ag(s)
Given : E⁰ for Zn²+/Zn=-0.76v,E⁰for Ag+/Ag=+0.80v
Given: log 10=1

Answer 1

Answer:

Here Zn is in the anode and Cu is in the cathode

Ecell =Ecathode-Eanode

therefore ;Ecell=0.34-(-0.76)

=1.10 V

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