Chromium plating is applied by electrolysis to objects suspended in a dichlromate solution , according to following `(` unbalanced `)` hald reaction `:` `Cr_(2)O_(7)^(2-)(aq) +e^(-) +H^(o+)(aq) rarr Cr(s)+H_(2)O(l)` How many hours would it take to apply a chromium plating of thickness `2.0xx10^(-2)mm` to a car bumper of suface area `0.25m^(2)` in an electrolysis cell carrying a current of `75.0A?` `[` Density of chromium is `7.19g cm^(-3)]`A. `2.2h`B. `1.5h`C. `3.0h`D. `0.25h`

Chromium plating is applied by electrolysis to objects suspended in a

1.	Chromium plating is applied by electrolysis to objects suspended in a dichlromate solution , according to following `(` unbalanced `)` hald reaction `:` `Cr_(2)O_(7)^(2-)(aq) +e^(-) +H^(o+)(aq) rarr Cr(s)+H_(2)O(l)` How many hours would it take to apply a chromium plating of thickness `2.0xx10^(-2)mm` to a car bumper of suface area `0.25m^(2)` in an electrolysis cell carrying a current of `75.0A?` `[` Density of chromium is `7.19g cm^(-3)]`A. `2.2h`B. `1.5h`C. `3.0h`D. `0.25h`
Answer» Correct Answer - b Volume to be covered `=(0.25xx10^(4)cm^(2))xx(2xx10^(-3)cm)=5cm^(3)` `implies` Mass of `Cr` to be deposited `=5xx7.19g` `=35.95g` `-=(35.95)/(52)mol` `=0.69mol` `Cr_(2)O_(7)^(2-)+14H^(o+) +12e^(-) rarr 2Cr+7H_(2)O` Now, `2 mol Cr` deposited `-=12 F` electricity `implies 0.59 mol Cr` deposited `-=(12)/(2)xx0.069F` `=(It)/(96500)=(75xxt)/(96500)t=1.48h~~1.5h`

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 ?

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment