Home
Class 11
CHEMISTRY
When a solution of AgNO3 (1 M) is electr...

When a solution of `AgNO_3` (1 M) is electrolysed using platinum anode and copper cathode, what are the products obtained at two electrodes?
Given :`E_(cu^(2+)|cu)^(@)=+0.34 "volt", E_(O_2,H^+|H_2O)^(@)=+1.23"volt",E_(H^+|H_2)^(@)=+0.0"volt" E_(ag^+|Ag)^(@)=+0.8 "volt"`

A

`CutoCu^(2+)` at anode, `Ag^+to Ag` at cathode

B

`H_2OtoO_2` at anode, `Cu^(2+)to Cu` at cathode

C

`H_2OtoO_2` at anode, `Ag^+to Ag` at cathode

D

`NO_3^-to NO_2` at anode, `Ag^+to Ag` at cathode

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 1 (Q.1 To Q.30)|1 Videos
  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 1 (Q.91 To Q.120)|1 Videos
  • DILUTE SOLUTION

    NARENDRA AWASTHI|Exercise Level 3 - Match The Column|1 Videos
  • GASEOUS STATE

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos

Similar Questions

Explore conceptually related problems

Which of the following statements is correct ? If E_(Cu^(2+)|Cu)^(@) = 0.34 V and E_(Sn^(2+)|Sn)^(@) = -0.136 V , E_(H^(+)|H_2)^(@) = -0.0V

The product obtained at anode when 50% H_(2)SO_(4) aqueous solution is electrolysed using platinum electrodes, is

A 0.200M KOH solutions is electrolysed for 1.5 h using a current of 8.00 A. How many moles O_2 were produced at the anode?

Calculate the Electrode potential of single electrode. Cu^(2+)(0.01M)//Cu" "(E^(@)= +0.337V)

Cu^+ ion is not stable in aqueous solution because of diproportionation reaction . E^(@) value for the disproportionation of Cu^(+) is : (E_(cu^(2+)// cu^+)^(@) = + 0.15 V , E_(cu^(+) // cu)^(@)= + 0.34 V )

An aqueous solution containing one mole per litre of each Cu(NO_3)_2 , AgNO_3 . Hg(NO_3)_(2), Mg(NO_3)_2 is being electrolysed using inert electrodes. The values of standard electrode potential (reduction potential) in volts are Ag//Ag^(+) = + 0.80 V " "Hg//Hg^(2+) = + 0.79V " " Cu//Cu^(2+) = +0.34 V " " Mg//Mg^(2+) = -2.37 V With increasing voltage, the sequence of deposition of metals on cathode will be

Copper reduces NO_3^(-) into NO and NO_2 depending upon concentration of HNO_3 in solution . Assuming [Cu^(2+)] = 0.1 M , and P_(NO) = P_(NO_2) = 10^(-3) bar . At which concentration of HNO_3 . Thermodynamic tendency for reduction of NO_3^(-) into NO and NO_2 by copper is same ? Given : E_(Cu^(2+)|Cu)^(@) = + 0.34 V , E_(NO_3^(-)|NO)^(@) = + 0.96 V , E_(NO_3^(-)|NO_2)^(@) = + 0.79 V

The emf (in V) of a Daniell cell containing 0.1 M ZnSO_(4) and 0.01 M CuSO_(4) solutions at their respective electrodes is (E_((Cu^(2+))/(Cu))^@ = +0.34 V, E_((Zn^(2+))/(Zn))^@ = -0.76 V)