According to the half-reaction table, `Sn^(2+) + 2e^(-) rightarrow Sn` `E_(@) = -0.14V` `Mn^(2+) + 2e^(-) rightarrow Mn` `E^(@) = -1.03V` Which species in the better oxidizing agent?A. `Mn^(2+)`B. `Sn^(2+)`C. MnD. Sn

According to the half-reaction table, `Sn^(2+) + 2e^(-) rightarrow Sn`

1.	According to the half-reaction table, `Sn^(2+) + 2e^(-) rightarrow Sn` `E_(@) = -0.14V` `Mn^(2+) + 2e^(-) rightarrow Mn` `E^(@) = -1.03V` Which species in the better oxidizing agent?A. `Mn^(2+)`B. `Sn^(2+)`C. MnD. Sn
Answer» Correct Answer - B

Discussion

No Comment Found

Related InterviewSolutions

The EMF of the following cell: `Zn| Zn^(2+)(0.1M) || Cl^(-)(0.01M) | AgCl | Ag` is: [Given : E_(Zn^(2+)|/Zn)^(@) = -0.76V`, `E_(Cl^(-)| AgCl|Ag)^(@)` = 0.22 V`, `(2.303RT)/(F) = 0.06]`A. 0.96 VB. 0.345 VC. 1.23D. 1.13 V
`DeltaG^(@)` of the cell reaction `AgCl(s) + 1/2H_(2) rightarrow Ag(s) + H^(+) + Cl^(-)` is `-21.52kJ`. `DeltaG^(@)` of 2AgCl(s) + H_(2)(g) rightarrow + 2H^(+) + 2Cl^(-)` isA. `-21.52kJ`B. `-10.76 kJ`C. `-43.04kJ`D. `43.04 kJ`
The emf of the cell `Zn|ZnCl_(2)` (0.05M) | AgCl(s)`,Ag is 1.015 V at 298 K and the temperature coefficient of its emf is `-4.92 xx 10^(-4)` V/K. How many of the reaction thermodynamic parameter `DeltaG`, `DeltaS` and `DeltaH` are negative at 298 K?A. None of themB. One of themC. Two of themD. All of them
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. 4.5 VB. 3.0 VC. 2.5 VD. 5.0 V
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^(-)`
For the electrolytic production of `NaClO_(4)` from `NaClO_(3)` as per the following equation: `NaClO_(3) + H_(2)O rightarrow NaClO_(4) + H_(2)` How many faradays of electricity will be required to produce 0.5 "mole" of `NaClO_(4)` assuming 60% efficiency?A. 0.835 FB. 1.67 FC. 3.34 FD. 1.6 F
What must be concentration of `Ag^(+)` in an aqueous solution containing `Cu^(2+) =1.0 M` so that bot the metals can be deposited on the cathode simultaneously. Given that `E_(Cu//Cu^(2+))^(0)=-0.34V` and `E_(Ag^(+)//Ag)^(0)=0.812V,T=298K`A. Nearly `10^(-19)` MB. `10^(-12)` MC. `10^(-8)` MD. Nearly `10^(-16)` M
Which process occurs in the electrolysis of an aqueous tin (II) chloride solution at a tin anode?A. Sn = `Sn^(2+) + 2e^(-)`B. `2Cl^(-) = Cl_(2) + 2e^(-)`C. `2H_(2)O = O_(2) + 4H^(+) + 4e^(-)`D. None of these
How many electrons flow when a current of 5 amperes is passed through a conductor for 200 seconds?A. `6.241 xx 10^(21)`B. `6.0241 xx 10^(21)`C. `6.241 xx 10^(22)`D. `6.0241 xx 10^(20)`
The number of electrons delivered at the cathode during electrolysis by a current of `1` ampere in 60 seconds is (charger on electron `= 1.60 xx 10^(-19)C`)A. `3.74 xx 10^(20)`B. `6.0 xx 10^(23)`C. `7.48 xx 10^(21)`D. `6.0 xx 10^(20)`

According to the half-reaction table, `Sn^(2+) + 2e^(-) rightarrow Sn` `E_(@) = -0.14V` `Mn^(2+) + 2e^(-) rightarrow Mn` `E^(@) = -1.03V` Which species in the better oxidizing agent?A. `Mn^(2+)`B. `Sn^(2+)`C. MnD. Sn

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment