A cell is containing two H electrodes. The negative electrode is in contact with a solution of `10^(-6)MH^(+)` ion. The e.m.f. of the cell is `0.118` volt at `25^(@)C` . Calculate `[H^(+)]` at positive electrode.

A cell is containing two H electrodes. The negative electrode is in co

1.	A cell is containing two H electrodes. The negative electrode is in contact with a solution of `10^(-6)MH^(+)` ion. The e.m.f. of the cell is `0.118` volt at `25^(@)C` . Calculate `[H^(+)]` at positive electrode.
Answer» Anode: `H_(2)rarr2H^(+)2e` (nagative polarity) `[H^(+)]=10^(6)M` `"Cathode: "2H^(+)+2erarrH_(2)` (positive polarity) `[H^(+)]rarraM` `thereforeE_(cell) =E_(OP_(H//H^(+)))+E_(Rp_(H^(+)//H))` `=E_(OP_(H//H^(+)))^(@) (0.059)/(2) log_(10)[H^(+)]_("anode")^(2)+E_(RP_(H^(+)//H))^(0)+(0.059)/(2)log_(10) [H^(+)]_("cathode")^(2)` `=(0.059)/(2)log_(10).([H^(+)]_("cathode")^(2))/([H^(+)]_("anode")^(-2))` `0.118=(0.059)/(2)log_(10).([H^(+)]_("cathode")^(2))/((10^(-6))^(2))` `=(0.059)/(2)log_(10).([H^(+)]_("cathode")^(2))/(10^(-6))` `therefore[H^(+)]_("cathode")=10^(-6)M`

Discussion

No Comment Found

Related InterviewSolutions

An alloy of Pb-Ag weighing `1.08g` was dissolved in dilute `HNO_(3)` and the volume made to 100 mL.A ? Silver electrode was dipped in the solution and the emf of the cell dipped in the solution and the emf of the cell set-up as `Pt(s),H_(2)(g)|H^(+)(1M)||Ag^(+)(aq.)|Ag(s)` was `0.62 V` . If `E_("cell")^(@)` is `0.80 V`, what is the percentage of Ag in the alloy ? (At `25^(@)C, RT//F=0.06`)A. 25B. `2.50`C. 10D. 50
Iron rod is dipped in concentrated `HNO_(3)` After some time the iron rod dipped in `AgNO_(3)` solution. Ag is not displace by Iron. This is becauseA. SRP of silver is less than ironB. Iron and silver have same lattice structureC. Iron becomes passiveD. All the above
For the electrolytic production of `NaClO_(4)` from `NaClO_(3)` according to the reaction `NaClO_(3)+H_(2)O rarr NaClO_(4)+H_(2). ` How many faradays of electricity would be required to produce `0.5 mo l e` of `NaClO_(4)`?A. 1B. 2C. 3D. `1.5`
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple mode for such a concentration cell involving a metal `M` is : `M_((s))|M_((aq.))^(@) 0.05"molar"||M_((aq.))^(@) 1"molar"|M_((s))` For the above electrolytic cell the magnitude of the cell potential `|E_(cell) | = 70mV` For the above cell :A. `E_(cell)lt0,DeltaGgt0`B. `E_(cell)gt0,Deltalt0`C. `E_(cell)lt0,DeltaH^(@)gt0`D. `E_(cell)gt0,DeltaG^(@)gt0`
Zinc reacts with `CuSO_(4)` according to the equation `Zn+CuSO_(4)rarrZnSO_(4)+Cu` . If excees of zinc is added to `100.0` ml of `0.05` M `CuSO_(4)` the amount of copper formed in moles will beA. `0.05`B. `0.5`C. `0.005`D. `50`
Aqueous solution of `NaCl` containing a small amount of `MeOH` ( methyl orange ) is electrolysed using `Pt-` electrodes . The color of the solution after some time will .A. remains yellowB. change from yellow to colorlessC. change from yellow to redD. remain red
List-I A) Electrolysis of aq. `Na_(2)SO_(4)` using Pt electrodes B) The charge carried by `6.023xx10^(23)` electrons is C) The amount of electricity required to deposit 27 grmams of Aluminium at cathode from molten `Al_(2)O_(3)` is D) A gas in contact with an inert electrode. List-II 1) 1 Faraday 2) 3 Faradays 3) `H_(2(g))//pt` 4) `O_(2)` at anode `H_(2)` at cathodeA. `{:(A,B,C,D),(2,3,4,1):}`B. `{:(A,B,C,D),(4,1,2,3):}`C. `{:(A,B,C,D),(3,2,4,1):}`D. `{:(A,B,C,D),(4,3,2,1):}`
The cell in whiCHM the following reaction occurs `:` `2Fe^(3+)(aq)+2I^(c-)(aq) rarr 2Fe^(2+)(aq)+I_(2)(s)` has `E^(c-)``_(cell)=0.2136V` at `298K`. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
`wedge^(@)._(m)` for `CaCl_(2)` and `MgSO_(4)` from the given data. `lambda_(Ca^(2+))^(@)=119.0S cm^(2)mol^(-1)` ltbr. `lambda_(Cl^(c-))^(@)=76.3S cm^(2)mol^(-1)` `lambda_(Mg^(2+))^(@)=106.0S cm^(2)mol^(-1)` `lambda_(SO_(4)^(2-))^(@)=160.0 cm^(2)mol^(-1)`
The molar conductivity of `0.25 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Calculate the degree of dissociation constant. Given `: lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) ` and `lambda_((CHM_(3)COO^(c-)))^(@)=54.6Scm^(2)mol^(-1)`

A cell is containing two H electrodes. The negative electrode is in contact with a solution of `10^(-6)MH^(+)` ion. The e.m.f. of the cell is `0.118` volt at `25^(@)C` . Calculate `[H^(+)]` at positive electrode.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment