When `E_(Ag)^(@)+_(//Ag)=0.8V` and ` E_(Zn^(@))2^(+)._(//Zn)` `=-0.76V`. Which of the following is correct?A. `Ag^(+)` can be reduced by `H_(2)`B. `Ag` can oxidise `H_(2)` into `H^(+)`C. `Zn^(2+)` can be reduced by `H_(2)`D. Ag can reduce `Zn^(2+)` ion.

When `E_(Ag)^(@)+_(//Ag)=0.8V` and ` E_(Zn^(@))2^(+)._(//Zn)` `=-0.76V

1.	When `E_(Ag)^(@)+_(//Ag)=0.8V` and ` E_(Zn^(@))2^(+)._(//Zn)` `=-0.76V`. Which of the following is correct?A. `Ag^(+)` can be reduced by `H_(2)`B. `Ag` can oxidise `H_(2)` into `H^(+)`C. `Zn^(2+)` can be reduced by `H_(2)`D. Ag can reduce `Zn^(2+)` ion.
Answer» Correct Answer - A Zinc has lower red pot, than hydrogen and hence can be reduced. Ag has more reduction potential so it is reduced easily.

Discussion

No Comment Found

Related InterviewSolutions

Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of a solution for a weak and a strong electrolyte.
Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:A. `Lambda_(m (NH_(4)Cl ))^(@) + Lambda_(m (NaCl))^(@) - Lambda_(m(NaOH))^(@)`B. `Lambda_(m (NaOH))^(@) + Lambda_(m(NaCl))^(@) - Lambda_(m(NH_(4)Cl))^(@)`C. `Lambda_(m(NH_(4)OH))^(@) + Lambda_(m (NH_(4)Cl))^(@) - Lambda_(m(HCl))^(@)`D. `Lambda_(m (NH_(4)Cl))^(@) + Lambda_(m (NaOH))^(@) - Lambda_(m(NaCl))^(@)`
Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:A. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(Na_(4)Cl)-Lambda_(m)^(@)(NaOH)`B. `Lambda_(m)^(@)(NaOH)+Lambda_(m)^(@)(NaCl)-Lambda_(m)^(@)(NH_(4)Cl)`C. `Lambda_(m)^(@)(NH_(4)OH)+Lambda_(m)^(@)(NH_(4)Cl)-Lambda_(m)^(@)(HCl)`D. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_(m)^(@)(NaCl)`
For the cell reaction `Cu^(2+)(C_(1),aq.)+Zn(s)hArrZn^(2+)(C_(2),aq)+Cu(s)` of an electrochemical cell, the change in free energy `(DeltaG)` of a given temperature is a function ofA. `lnC_(1)`B. `lnC_(2)//C_(1)`C. `ln (C_(1) + C_(2))`D. `ln C_(2)`
Use of lithium metal as an electrode in high energy density batteries is due to:A. Lithium is highest elementB. Lithium has highest oxidation potentialC. Lithium is quite reactiveD. Lithium does not corrode readily
The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follow : `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2), Delta_(r)G= +960 kJ mol^(-1)` The potential difference needed for the electrolytic reduction of aluminium oxide `(Al_(2)O_(3))` at `500^(@)C` isA. `5.0 V`B. `4.5 V`C. `3.0 V`D. `2.5 V`
Based on the data given below, the correct order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt A1`B. `Fe^(2+) lt Al lt Br^(-)`C. `Al^(-) lt Br^(-) lt Fe^(2+)`D. `Al^(-) lt Fe^(2+) lt Br^(-)`
Based on the data given below, the correcy order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt Al`B. `Fe^(2+) lt Al lt Br^(-)`C. `Al lt Br^(-) lt Fe^(2+)`D. `Al lt Fe^(2+) lt Br^(-)`
Based on the data given below, the correcy order of reducing power is: `Fe_((aq.))^(3+) + e rarr Fe_((aq.))^(2+), E^(@) = +0.77 V` `Al_((aq.))^(3+) + 3e rarr Al_((s)), E^(@) = -1.66 V` `Br_(2(aq.)) + 2e rarr 2Br_((aq.))^(-), E^(@) = +1.08 V`A. `Br^(-) lt Fe^(2+) lt Al`B. `Fe^(2+) lt Al lt Br^-`C. `Al lt Br^(-) lt Fe^(2+)`D. `Al lt Fe^(2+) lt Br^(-)`
What are elementary reactions ?

When `E_(Ag)^(@)+_(//Ag)=0.8V` and ` E_(Zn^(@))2^(+)._(//Zn)` `=-0.76V`. Which of the following is correct?A. `Ag^(+)` can be reduced by `H_(2)`B. `Ag` can oxidise `H_(2)` into `H^(+)`C. `Zn^(2+)` can be reduced by `H_(2)`D. Ag can reduce `Zn^(2+)` ion.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment