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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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The correct Answer is:
`-0.2205V`
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Calculate the standard electrode potential of the Ni^(2+)//Ni electrode , if the cell potential potential of the cell, Ni//N^(2+)(0.01 M)//Cu is 0.59" V ". "Given" E_(Cu^(2+)//Cu)^(@)=+0.34 " V "

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+) ?

Emf of the cell Ni| Ni^(2+) ( 0.1 M) | Au^(3+) (1.0M) Au will be E_(Ni//Ni(2+))^@ = 0.5=25. E_(Au//Au^(3+))^@ = 1.5 V .

(a) A cell is prepared by dipping a zinc rod in 1M zinc sulphate solution and a silver electrode in 1M silver nitrate solution. The standard electrode potential given : E^(@)Zn_(2+1//Zn) = -0.76V, E^(@)A_(g+//)A_(g) = +0.80V What is the effect of increase in concentration of Zn^(2+) " on the " E_(cell) ? (b) Write the products of electrolysis of aqueous solution of NaCI with platinum electrodes. (c) Calculate e.m.f. of the following cell at 298 K: "Ni(s)"//"Ni"^(2+)(0.01M)////"Cu"^(2+)(0.1M)//"Cu(s)" ["Given"E_(Ni2+//Ni)^(@) = -0.025 V E_(Cu2+//Cu)^(@) = +0.34V] Write the overall cell reaction.

If standard electrode potenial of Cu^(2+)//Cu is 0.34V then potential of Cu dipped in 0.1 M solution of CuSO_(4) will be

Calculate the e.m.f of the cell Mg(s)//Mg^(2+)(0.1 M)||Cu^(2+)(1.0xx10^(-3) M)//Cu(s) Given E_(Cu^(2+)//Cu)^(@)=+0.34 V and E_(Mg^(2+)//Mg)^(@)=-2.37 V

PRADEEP-ELECTROCHEMISTRY-Problem for Practice
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