(a) Calculate the potential of `Zn^(2+)//Zn` electrode in which zinc ion activity is 0.001 M. `mol^(-1),=96500 C mol^(-1), E_(Zn^(2+)//Zn)^(@)=-0.76"

Question

(a) Calculate the potential of `Zn^(2+)//Zn` electrode in which zinc ion activity is 0.001 M. `mol^(-1),=96500 C mol^(-1), E_(Zn^(2+)//Zn)^(@)=-0.76"

(a) Calculate the potential of `Zn^(2+)//Zn` electrode in which zinc ion activity is 0.001 M. `mol^(-1),=96500 C mol^(-1), E_(Zn^(2+)//Zn)^(@)=-0.76" V ")`
(b) Calculate the electrode potential of copper electrode dipped in a 0.1 M solution of copper sulphate at 298 K, assuming `CuSO_(4)` to be completely ionised. The standard reduction potential of copper , `E_(Cu^(2+)//Cu)^(@)=0.34" V "` at 298 K.

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - 0.31045 V
(a) Electrode reaction : `Zn^(2+)(0.001" M") +2e^(-) to Zn(s)`
`E_((Zn^(2+)//Zn))-E_((Zn^(2+)//Zn))^(@)-(0.0591)/(n)"log"(1)/([Zn^(2+)])=-0.76-(0.0591)/(2)"log"(1)/([10^(-3)])`
`=-0.34-0.02955xx3=-0.76-0.08865=-0.84865" V "`
(b) Electrode reaction : `Cu^(2+)(0.1" M")+2e^(-) to Cu(s)`
`E_((Cu^(2+)//Cu))-E_((Cu^(2+)//Cu))^(@)-(0.0591)/(2)"log"(1)/([Cu^(2+)])=-0.34-0.02955" log"(1)/((0.1))`
`=0.34-0.02955=0.31045" V "`