Home
Class 12
CHEMISTRY
One half-cell in a voltaic cell is const...

One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration . Its other half-cell consists of a zinc electrode dipping in `1.0`M solution of `Zn(NO_(3))_(2)` . A voltage of `1.48` V is measured for this cell . Use this information to calculate the concentration of silver nitrate solution used. `(E_(Zn^(2+)|Zn)^(@) = -0.76V , E_(Ag^(+)|Ag)^(@) = + 0.80V`).

Text Solution

AI Generated Solution

To solve the problem, we will follow these steps: ### Step 1: Identify the half-reactions In the voltaic cell, we have two half-reactions: - **Cathode (Reduction)**: Silver ions are reduced to solid silver. \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (E^\circ = +0.80 \, \text{V}) \] ...
Promotional Banner

Topper's Solved these Questions

  • XII BOARDS

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise [SET-I]|71 Videos

  • XII BOARDS

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise [SET-II]|10 Videos

  • XII BOARDS

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise Delhi Board : Set - II|7 Videos

  • SAMPLE PAPER 2019

    XII BOARD PREVIOUS YEAR PAPER ENGLISH|Exercise SECTION: D|1 Videos

Similar Questions

Explore conceptually related problems

Explain the following : Zinc displaces hydrogen from acid solution. E_(Zn^(2+)//Zn)^(Theta)= -0.76V

Calculate the potential of silver electrode in a saturated solution of AgBr(K_(sp)=6xx10^(-13)) containing 0.1 M KB r . E^(c-)._(Ag^(o+)|Ag)=0.80V.

Can a solution of 1 M ZnSO_(4) be stored in a vessel made of copper ? Given that E_(Zn^(+2)//Zn)^(@) =-0.76V and E_(Cu^(+2)//Cu)^(@)=0.34 V

A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ions is not known.The cell potential measured is 0.422 V. Determine the concentration of silver ions in the cell. [Given E_(Ag^(+)//Ag)^(@)=0.80,E_(Cu^(2+)//Cu)^(@)=+0.34 V]

What is the effect of increase in concentration of zinc ions on the electrode potential of zinc electrode for which E_(Zn^(2+)//Zn) equals - 0.76 V

A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn (NO_(3))_(2) solution and metallic plate of lead in 0.02M Pb(NO_(3))_(2) solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. ("Given" : E^(@) Zn^(2+)ZN=0.76V, " " E^(@)Pb^(2+)//Pb = -0.13V)

Find the E_(cell)^(@) for the following cell reaction Fe^(+2)+Zn rarr Zn^(+2)+Fe Given E_(Zn//Zn^(+2))^(@)=0.76V,E_(Fe//Fe^(+2))^(@)=+0.41 V

If excess of Zn is added to 1.0M solution of CuSO_(4) , find the concentration of Cu^(2+) ions at equilibrium. Given :E^(c-)._((Zn^(2+)|Zn))=-0.76V E^(c-)._(cell)=(E^(c-)._((Cu^(2+)|Cu))=0.34V

A silver electrode dipping in AgNO_(3) solution (0.1M) is combined salt bridge with a hydrogen electrode dipping in a solution of pH=3(at 25^(@)C) . If the standard reduction potential of the silver electrode is 0.799V , what is the EMF of the cell ?

Write the cell reaction and calculate the standard E^(0) of the cell: Zn | Zn^(2+)(1M) | | Cd^(2+)(1M) | Cd Given E_(Zn|Zn^(2+))^(0)= 0.763 volt E_(Cd|Cd^(2+))^(0) = 0.403 volt