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Calculate the standard electrode potenti...

Calculate the standard electrode potential of `Ni^(2+)`/Ni electrode if emf of the cell `Ni_((s)) |Ni^(2+) (0.01M)| |CU^(2)| Cu _((s))(0.1M)` is `0.059 V.` `[Given : E_(Cu^(2+)//Cu)^(@) =+0.34V`

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To calculate the standard electrode potential of the `Ni^(2+)`/Ni electrode, we will follow these steps: ### Step 1: Write the cell reaction The cell reaction involves nickel and copper: \[ \text{Ni (s)} \rightarrow \text{Ni}^{2+} + 2e^- \] \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu (s)} \] ### Step 2: Identify the anode and cathode ...
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If the standard electrode poten tial of Cu^(2+)//Cu electrode is 0.34V. What is the electrode potential of 0.01 M concentration of Cu^(2+) ?

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Calculate the e.m.f. of the cell Zn//Zn^(2+) (0.1M) "||" Cu^(2+) (0.01M) "|"Cu E_(Zn^(2+)//Zn)^(Theta)= -0.76 " V and "E_(Cu^(2+)//Cu)^(Theta)=0.34V

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Calculate the emf of the given cell: Cu/ Cu^(2+)(0.01M) ∣∣ Cu^(2+) (0.001M)∣Cu is ( E^(@)Cu^(2+) /Cu=+0.34V)

E_(Cu^(2+)//Cu)^(@) =0.43V E_(Cu^(+)//Cu)^(@)=0.55V E_(Cu^(2+)//Cu^(+))^(@) =

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

E_(Cu^(2+)//Cu)^(@)=0.34V E_(Cu^(+)//Cu)^(@)=0.522V E_(Cu^(2+)//Cu^(+))^(@)=

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